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George Mason University General Chemistry 212 Chapter 19 Ionic Equilibria Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Change, 6 th ed, 2011, Martin S. Silberberg, McGraw-Hill - PowerPoint PPT Presentation
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1/23/2015 1
George Mason UniversityGeneral Chemistry 212
Chapter 19Ionic Equilibria
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7 Change, 7thth edition, 2011, McGraw-Hill edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.
1/23/2015 2
Ionic Equilibria in Aqueous Solutions
Equilibria of Acid-Base Buffer Systems The Common Ion Effect The Henderson-Hasselbalch Equation Buffer Capacity and Range Preparing a buffer
Acid-Base Titration Curves Acid-Base Indicators Strong-Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Weak Base-Strong Acid Titrations Polyprotic Acid Titrations Amino Acids as Polyprotic Acids
1/23/2015 3
Ionic Equilibria in Aqueous Solutions
Equilibria of Slightly Soluble Ionic Compounds
The Solubility-Product Constant Calculations involving Ksp
The Effect of a Common Ion
The Effect of pH Qsp vs. Ksp
Equilibria involving Complex Ions
Formation of Complex Ions
Complex Ions and Solubility
Amphoteric Hydroxides
1/23/2015 4
Ionic Equilibria in Aqueous Solutions
The simplest acid-base equilibria are those in which a single acid or base solute reacts with water
In this chapter, we will look at:
solutions of weak acids and bases through acid/base ionization
The reactions of salts with water
Titration curves
All of these processes involve equilibrium theory
1/23/2015 5
Ionic Equilibria in Aqueous Solutions
Equilibria of Acid-Base Buffer Systems
Buffer
An Acid-Base Buffer is a species added to a solution to minimize the impact on pH from the addition of [H3O+] or [OH-] ions
Small amounts of acid or base added to an unbuffered solution can change the pH by several units
Since pH is a logarithmic term, the change in [H3O+] or [OH-] can be several orders of magnitude
1/23/2015 6
Ionic Equilibria in Aqueous Solutions
The Common Ion Effect
Buffers work through a phenomenon known as the:
Common Ion Effect
The common ion effect occurs when a given ion is added to an “equilibrium mixture of a weak acid or weak base that already contains that ion
The additional “common ion” shifts the equilibrium away from its formation to more of the undissociated form; i.e., the acid or base dissociation decreases
A buffer must contain an “acidic” component that can react with the added OH- ion, and a “basic” component than can react with the added [H3O+]
1/23/2015 7
Ionic Equilibria in Aqueous Solutions
The buffer components cannot be just any acid or base
The components of a buffer are usually the conjugate acid-base pair of the weak acid (or base) being buffered
Ex: Acetic Acid & Sodium Acetate
Acetic acid is a weak acid, slightly dissociated
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Acid Base Conjugate Conjugate
Base Acid
1/23/2015 8
Ionic Equilibria in Aqueous Solutions
Ex: Acetic Acid & Sodium Acetate (Con’t)
Now add Sodium Acetate (CH3COONa), a strong electrolyte (Acetate ion is the conjugate base)
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) (added)
The added Acetate ions shift the equilibrium to the left forming more undissociated acetic acid
This lowers the extent of acid dissociation, which lowers the acidity by reducing the [H3O+] and increasing the pH
Similarly, if Acetic Acid is added to a solution of Sodium Acetate, the Acetate ions already present act to suppress the dissociation of the acid
1/23/2015 9
Ionic Equilibria in Aqueous Solutions
When a small amount of [H3O+] is add to acetic acid/acetate buffer, that same amount of acetate ion (CH3COO-) combines with it, increasing the concentration of acetic acid (CH3COOH).
The change in the [HA]/[A-] ratio is small; the added [H3O+] is
effectively tied up; thus, the change in pH is also small
1/23/2015 10
Ionic Equilibria in Aqueous Solutions
Essential Features of a Buffer
A buffer consists of high concentrations of the undissociated acidic component [HA] and the conjugate base [A-] component
When small amounts of [H3O+] or [OH-] are added to the buffer, they cause small amounts of one buffer component to convert to the other component
1/23/2015 11
Ionic Equilibria in Aqueous Solutions
Ex. When a small amount of H3O+ is added to an Acetate buffer, that same amount of CH3COO- combines with it, increasing the amount of undissociated CH3COOH tying up potential [H3O+]
Similarly, a small amount of OH- added combines with undissociated CH3COOH to form CH3COO- & H2O tying up potential [OH-]
In both cases, the relative changes in the amount of buffer components is small, but the added [H3O+] or [OH-] ions are tied up as undissociated components; thus little impact on pH
1/23/2015 12
Ionic Equilibria in Aqueous Solutions
The equilibrium perspective
The [H3O+] (pH) of the solution depends directly on the buffer-component concentration ratio
If the ratio [HA]/[A-] goes up, [H3O+] goes up (pH down)
If the ratio [HA]/[A-] goes down, [H3O+] goes down([OH-] goes up) and pH increases (less acidic)
When a small amount of a strong acid is added, the increased amount of [H3O+] reacts with a stoichiometric amount of acetate ion from the buffer to form more undissociated acetic acid
+ -+ -3 3
a3
[H O ][CH COO ][H ][A ]K = =
HA [CH COOH]
+ 33 a a --
3
HA [CH COOH][H O ] = K × = K ×
[CH COO ]A
+ -3 2 3 3CH COOH + H O = H O + CH COO
1/23/2015 13
Practice ProblemWhat is the pH of a buffer solution consisting of 0.50 M CH3COOH (HAc) and 0.50 M CH3COONa (NaAc)?
Ka (CH3COOH) = 1.8 x 10-5
+3 2[H O ] from autoionization of H O is neglible and is neglected
Concentration HAc(aq) + H2O(l) Ac-(aq) + H3O+
Initial Ci 0.50 - 0.50 0
- X - + X +X
Equilibrium Cf 0.50 - X - 0.50 + X X
If x is small, i.e., < 5% CH3COOHinit , then:
3[CH COOH] = 0.50 - x 0.50-
3[CH COO ] = 0.50 + x 0.50
+3 dissoc 3x = [CH COOH] = [H O ]
(Con’t)
NaAc is a sodium salt : it dissociates completely
1/23/2015 14
Practice Problem (Con’t)
+ 33 a -
3
[CH COOH]x = [H O ] = K ×
[CH COO ]
-+ -33 3
a3 3
x ×[CH COO ][H O ][CH COO ]K = =
[CH COOH] [CH COOH]
+ -5 -53
0.5x = [H O ] = 1.8 10 × = 1.8×10 M
0.5
+ -53pH = - log[H O ] = - log(1.8 10 )
pH = 4.74
1/23/2015 15
Practice Problem (Con’t)Calculate the pH of the previous buffer solution after adding 0.020 mol of solid NaOH to 1 L of the solution
Set up reaction table for the stoichiometry of NaOH (strong base) to Acetic acid (weak acid)
Concentration HAc(aq) + OH-(aq) Ac-(aq) + H2O(l)
Initial Ci 0.50 0.020 0.50 -
- 0.020 -0.020 + 0.020 -
Equilibrium Cf 0.48 0.0 0.52 -
NaOH0.020mol
Molarity (M) = = 0.020 mol / L1L
Set up reaction table for acid dissociation using newconcentrations
Concentration HAc(aq) + H2O Ac-(aq) + H3O+(aq)
Initial Ci 0.48 - 0.52 0
- X - + X + X
Equilibrium Cf 0.48 - X - 0.52 + X X(Con’t)
1/23/2015 16
Practice Problem (Con’t)Assuming x is small:
-3 3[CH COOH] = 0.48 M - x = 0.48 M and [CH COO ] = 0.52 M + x = 0.52 M
+ -533 a -
3
[CH COOH] 0.48x = [H O ] = K × = (1.8×10 ) ×
0.52[CH OO ]
+ -53[H O ] = 1.7 ×10 M
+ -53pH = -log[H O ] = -log(1.7 ×10 )
pH = 4.77
The addition of the strong base increased the
concentration of the basic buffer [CH3COO-] component to 0.52 M at the expense of the acidic buffer component [CH3COOH] (0.48 M)
The concentration of the [H3O+] (x) decreased, but
only slightly, thus, the pH increased only slightly from 4.74 to 4.77
1/23/2015 17
Ionic Equilibria in Aqueous Solutions
The Henderson-Hasselbalch Equation
For any weak acid, HA, the dissociation equation and Ka expression are:
The key variable that determines the [H3O+] is the ratio of acid species (HA) to base species (A-)
+ -2 3HA + H O H O + A
+ -3
a
[H O ][A ]K =
[HA]
+3 a -
[HA] (acid species)[H O ] = k ×
[A ] (base species)+
3 a -
[HA]-log[H O ] = - logK - log
[A ]-
a[A ]
pH = pK + log[HA]
a[base]
pH = pK + log[acid]
Note the change in sign of “log” term and inversion of acid-base terms
1/23/2015 18
Practice ProblemWhat is the pH of a buffer formed by adding 0.15 mol of acetic acid and 0.25 mol of sodium acetate to 1.5 L of water?
Ans: Both the acid and salt forms are added to the solution at somewhat equal concentrations
Neither form has much of a tendency to react
Dissociation of either form is opposed in bi-directional process
The final concentrations will approximately equal the initial concentrations[HA] = 0.15 mol / 1.5 L = 0.10 M
-[A ] = 0.25 mol / 1.5 L = 0.167 M-5
aK for acetic acid = 1.7 × 10 (from table "C")-5
a apK = - log(K ) = - log(1.7 10 ) = 4.77 (Con’t)
1/23/2015 19
Practice ProblemExample Problem (con’t)
From the Henderson-Hasselbalch equation-
a[A ]
pH = Pk + log[HA]
0.167MpH = 4.77 + log = 4.77 + log(1.67) = 4.77 + 0.22
0.10M
pH = 4.99
1/23/2015 20
Ionic Equilibria in Aqueous Solutions
Buffer Capacity A buffer resists a pH change as long as the
concentration of buffer components are large compared with the amount of strong acid or base added
Buffer Capacity is a measure of the ability to resist pH change
Buffer capacity depends on both the absolute and relative component concentrations Absolute - The more concentrated the components
of a buffer, the greater the buffer capacity Relative – For a given addition of acid or base, the
buffer-component concentration ratio changes less when the concentrations are similar than when they are different
1/23/2015 21
Ionic Equilibria in Aqueous Solutions
Ex. Consider 2 solutions
Solution 1 – Equal volumes of 1.0M HAc and 1.0 M Ac-
Solution 2 – Equal volumes of 0.1M HAc and 0.1M Ac-
Same pH (4.74) but 1.0 M buffer has much larger buffer capacity
(Con’t)
1/23/2015 22
Ionic Equilibria in Aqueous Solutions
Ex. Buffer #1 [HA] = [A-] = 1.000 M
Add 0.010 mol of OH- to 1.00 L of buffer
[HA] changes to 0.990 M [A-] changes to 1.010 M
-
init
init
[A ] 1.000 M = = 1.000
[HA] 1.000 M
-
final
final
[A ] 1.010 M = = 1.020
[HA] 0.990 M
1.020 -1.000Percent change = ×100 = 2%
1.000
1/23/2015 23
Ionic Equilibria in Aqueous Solutions
Buffer #2 [HA] = 0.250 M [A-] = 1.75 M
Add 0.010 mol of OH- to 1.00 L of buffer
[HA] changes to 0.240 M [A-] changes to 1.760 M
-
init
init
[A ] 1.750 M = = 7.000
[HA] 0.250 M
-
final
final
[A ] 1.760 M = = 7.330
[HA] 0.240 M
7.330 - 7.000Percent change = × 100 = 4.7%
7.000
Buffer-component concentration ratio is much larger when the initial concentrations of the components are very different
1/23/2015 24
Ionic Equilibria in Aqueous Solutions
A buffer has the highest capacity when the component acid / base concentrations are equal:
A buffer whose pH is equal to or near the pKa of its acid component has the highest buffer capacity
-
a a a a
[A ]pH = pK + log = pK + log1 = pK + 0 = pK
[HA]
-ABase = = 1
Acid HA
1/23/2015 25
Ionic Equilibria in Aqueous Solutions
Buffer Range pH range over which the buffer acts effectively is
related to the relative component concentrations The further the component concentration ratio is
from 1, the less effective the buffering action If the [A-]/[HA] ratio is greater than 10 or less
than 0.1 (or one component concentration is more than 10 times the other) the buffering action is poor
Buffers have a usable range within:
± 1 pH unit or pKa value of the acid components
a a
10pH = pK + log = pK + 1
1
a a
1pH = pK + log = pK - 1
10
1/23/2015 26
Ionic Equilibria in Aqueous Solutions
Equilibria of Acid-Base Systems
Preparing a Buffer
Choose the Conjugate Acid-Base pair
Driven by pH
Ratio of component concentrations close to 1 and pH ≈ pKa
Con’t
1/23/2015 27
Ionic Equilibria in Aqueous Solutions
Ex. Assume you need a biochemical buffer whose pH is 3.91. pKa of acid component should be close to 3.9
Ka = 10-3.9 = 1.3 x 10-4
2. From a table of pKa values select buffer possibilitiesLactic acid (pKa = 3.86)Glycolic acid (pKa = 3.83)Formic Acid (pKa = 3.74)
3. To avoid common biological species, select Formic Acid
Buffer components of Formic Acid
Formic Acid – HCOOH (Acid)
Formate Ion – HCOO- (Conjugate Base)
Obtain soluble Formate salt – HCOONa Con’t
1/23/2015 28
Ionic Equilibria in Aqueous Solutions
4. Calculate Ratio of Buffer Component Concentrations ([A-]/[HA]) that gives desired pH
1.4 mol of HCOONa is needed for every mole HCOOH
-[A ]pH = Pka + log
[HA]
-[HCOO ]log = 3.9 - 3.74 = 0.16
[HCOOH]
-[HCOO ]3.9 = 3.74 + log
[HCOOH]
Con’t
-0.16[HCOO ]
Thus : = 10 = 1.4[HCOOH]
1/23/2015 29
Ionic Equilibria in Aqueous Solutions
Preparing a Buffer (Con’t)
Determine the Buffer Concentrations
The higher the concentration of the components, the higher the buffer capacity
Assume 1 Liter of buffer is required and you have a stock of 0.40 M Formic Acid (HCOOH)
Compute moles and then grams of Sodium Formate (CHOONa) needed to produce 1.4/1.0 ratio
0.40 mol HCOOHMoles of HCOOH = 1.0 L × = 0.40 mol HCOOH
1.0 L soln
1.40 mol HCOONaMoles of HCOONa = 0.40 mol HCOOH × = 0.56 mol HCOONa
1.0 mol HCOOH
68.01 g HCOONaMass of HCOONa = 0.56 mol HCOONa × = 38 g HCOONa
1 mol HCOONa Con’t
1/23/2015 30
Ionic Equilibria in Aqueous Solutions
Preparing a Buffer (Con’t)
Mix the solution and adjust the pH
The prepared solution may not be an ideal solution (see Chapter 13, Section 6)
The desired pH (3.9) may not exactly match the actual value of the buffer solution
The pH of the buffer solution can be adjusted by a few tenths of a pH unit by adding strong acid or strong base.
1/23/2015 31
Practice ProblemExample of preparing a buffer solution without using the Henderson-Hasselbalch equationHow many grams of Sodium Carbonate (Na2CO3) must be added to 1.5 L of a 0.20 M Sodium Bicarbonate (NaHCO3) solution to make a buffer with pH = 10.0?
Ka (HCO3-) = 4.7 x 10-11 (From Appendix
C)
Con’t
Conjugate Pair - HCO3- (acid) and CO3
2- (base)
Convert pH to [H3O+]
- + 2-
3 2 3 3HCO (aq) + H O(l) H O (aq) + CO (aq)
-10+ -pH -10
3
[H O ] = 10 = 10 = 1×10 M
1/23/2015 32
Practice Problem (Con’t)3. Use Ka to compute [CO3
2-]
2-2- 2-3
3 30.094 mol CO
Moles CO = 1.5 L soln × = 0.14 mol CO1 L soln
5. Compute the mass (g) of Na2CO3 needed
2 32 3 2 3
2 3
105.99 g Na COMass (g) of Na CO = 0.14 mol Na CO × = 14.8 g
1mol Na CO
Con’t
4. Compute the amount of CO32- needed for 1.5 L of solution
2-- -113
3 a + -103
[HCO ] 0.20[CO ] = K = (4.7 10 ) = 0.094 M
[H O ] 1.0 10
+ 2-3 3
a -3
[H O ][CO ]K =
[HCO ]
1/23/2015 33
Practice Problem (Con’t)6. Dissolve the 14.8 g Na2CO3 in slightly less than 1.5 L of the
NaHCO3, say 1.3 L
7. Add additional NaHCO3 to make 1.5 L; adjust pH to 10.0with a strong acid
8. Check for a useful buffer range:
The concentration of the acidic component [HCO3-]
must be within a factor of “10” of the concentration of the basic component [CO3
2-]
- - -3 3 3HCO = 1.5 L × 0.20 M HCO = 0.30 mol HCO
- 2-3 30.30 mol HCO vs. 0.14 mol CO
0.30 / 0.14 = 2.1 << 10
Therefore, the buffer solution is reasonable
1/23/2015 34
Ionic Equilibria in Aqueous Solutions
Acid-Base Titration Curves Acid-Base Indicators
Weak organic acid (HIn) that has a different color than its conjugate base (In-)
The color change occurs over a relatively narrow pH range
Only small amounts of the indicator are needed; too little to affect the pH of the solution
The color range of typical indicators reflects a100 - fold range in the [HIn]/In-] ratio
This corresponds to a pH range of 2 pH units
1/23/2015 35
Mixing Acids & Bases Acids and bases react through neutralization reactions The change in pH of an acid mixed with a base is
tracked with an Acid-Base Titration Curve Titration:
Titration Curve: Plot of pH vs. the volume of the “Strong” acid or “Strong” base being added via buret
Solution in buret is called the “titrant” Equivalence Point: Point in a titration curve where
stoichiometric amounts of acid and base have been mixed (point of complete reaction)
3 Important Cases:
Strong Acid + Strong Base (& vice versa)
Weak Acid + Strong Base
Weak Base + Strong Acid
1/23/2015 36
Titration Curves Titration Curve: Plot of measured pH versus Volume
of acid or base added during a “Neutralization” experiment
All Titration Curves have a characteristic
“Sigmoid (S-shaped) profile
Beginning of Curve: pH changes slowly
Middle of Curve: pH changes very rapidly
End of Curve: pH changes very slowly again
pH changes very rapidly in the titration as the equivalence point (point of complete reaction) is reached and right after the equivalence point
1/23/2015 37
Neutralization Reactions Acids and Bases react with each other to form salts
(not always) and water (not always) through
Neutralization Reactions Neutralization reaction between a strong acid and
strong baseHNO3(aq) + KOH(aq) KNO3(aq) + H2O(l)
acid base salt water Neutralization of a strong acid and strong base
reaction lies very far to the right (Kn is very large); reaction goes to completion; net ionic equation is
H3O+(aq) + OH-(aq) 2 H2O(l)
H3O+ and OH- efficiently react with each other to form water
2142
3 3
n - -w
+ +[H O] 1 1
K = = = = 1×10K[H O ][OH ] [H O ][OH ]
1/23/2015 38
Titration Curves Curve for Titration of a Strong Acid by a Strong Base
1/23/2015 39
Practice Problem
Calculate pH after the addition of 10, 20, 30 mL NaOH during the titration of 20.0mL of 0.0010 M HCL(aq) with 0.0010 M NaOH(aq)
Initial pH HCl soln: [HCl] = [H3O+] = 0.0010 M
+ -3 2
Mixture
Strong Acid + Strong Base
H O (aq) + OH (aq) 2H O(l)€
+3pH = - log[H O ] = - log(0.0010) = 3.0
Con’t
1/23/2015 40
Practice ProblemAdd 10 mL 0.0010 M NaOH
1 mol H3O+ disappears for each mole OH- added
moles = Molarity(mol / L) × Volume(L)
+ +3 3moles of H O remaining = 0.00001 mol H O
+ +3 3Initial moles of H O = 0.0010 mol / L x 0.020L = 0.00002 mol H O
- -- moles OH added = 0.0010 mol / L x 0.010L = 0.00001 mol OH
Con’t
1/23/2015 41
Practice Problem (Con’t) Calculate [H3O+] & pH, taking “Total Volume” into
account
+
+ 33
0.000010 mol H O[H O ] = = 0.000033M
0.020L + 0.010L
+3pH = - log[H O ] = - log(0.0000333) = 3.48
Con’t
++ 3
3
amount(mol) of H O remaining[H O ] =
original volume of acid + volume of added base
1/23/2015 42
Practice Problem (Con’t) pH at “Equivalence Point”
Equal molar amounts of HCL & NaOH have been mixed
20 mL x 0.0010 M HCl reacts with20 mL x 0.0010 M NaOH
At the equivalence point all H3O+ has been neutralized by the addition of all the NaOH, except for [H3O+] from the autoionization of water, i.e., negligible
[H3O+] = pure water = 1 x 10-7 M
Con’t
1/23/2015 43
Practice Problem (Con’t) After the equivalence point
After the equivalence point, the pH calculations are based on the moles of excess OH- present
A total of 30 mL of NaOH has been added
- -moles of excess OH = 0.00001 mol OH
-[OH ] = 0.00020M
-pOH = - log[OH ] = - log(0.00020) = 3.70
wpH = pK - pOH = 14.0 - 3.7 = 10.3
- -- amount excess OH 0.000010 moles OH
[OH ] = =original volume acid + volume base added 0.020 L + 0.030 L
- -Total moles of OH added = 0.0010 mol / L x 0.030L = 0.00003 mol OH
+ +-3 3- moles H O consumed = 0.0010 mol / L x 0.020L = 0.00002 mol H O
1/23/2015 44
Neutralization Reactions Neutralization reactions between weak acids and
strong bases (net ionic equation shown below) C6H5COOH(aq) + NaOH C6H5COONa(aq) + H2O(l)
benzoic acid base benzoate “salt” water
The above equilibrium lies very far to the right The two equilibria below can be added together to
provide the overall neutralization reaction shown above
(1) C6H5COOH(aq) + H2O(l) C6H5COO-(aq) + H3O+
(aq)
Ka = 1.7 x 10-5 (from Appendix C)
(2) H3O+(aq) + OH-(aq) 2 H2O(l)
+ - -14w 3
14
w
Recall : K = [H O ][OH ] = 1.0 × 10
1 = 1.0×10K
1/23/2015 45
Neutralization Reactions The overall neutralization equilibrium
constant (Kn) is the product of the two intermediate equilibrium constants
(Ka & 1/Kw)
KN = Ka x 1/Kw = (1.7 x 10-5)(1 x 1014) = 1.9 x 109
(very large!)
Weak acids react completely with strong bases
1/23/2015 46
Titration CurvesCurve for Titration of a Weak Acid by a Strong Base
1/23/2015 47
Weak Acid-Strong Base Titration Curve
Consider the reaction between Propanoic Acid (weak acid) and Sodium Hydroxide (NaOH) (strong base)
Ka for CH3CH2COOH (HPr) - 1.3 x 10-5
The Titration Curve (see previous slide)
The bottom dotted curve corresponds to the strong acid-strong base titration
The Curve consists of 4 regions, the first 3 of which differ from the strong acid case
The initial pH is “Higher”
The weak acid dissociates only slightly producing less [H3O+] than with a strong acid
The gradually arising portion of the curve before the steep rise to the equivalence point is called the “buffer region”
1/23/2015 48
Weak Acid-Strong Base Titration Curve
As HPr reacts with the strong base, more and more of the conjugate base (Pr-) forms creating an (HPr/Pr-) buffer
At the midpoint of the “Buffer” region, half the original HPr has reacted
pH at midpoint of “Buffer” region is common method of estimating pKa of an “unknown” acid
-
- [Pr ]HPr = [Pr ] or = 1
[HPr]
-
a a a[Pr ]
pH = Pk + log = pK + log(1) = pK[HPr]
1/23/2015 49
Weak Acid-Strong Base Titration Curve
The pH at the “equivalence point” is greater than 7.00 The weak-acid anion (Pr-) acts as a
weak base to accept a proton from H2O forming OH-
The additional [OH-] raises the pH Beyond the equivalence point, the ph
increases slowly as excess OH- is added
1/23/2015 50
Practice ProblemCalculate the pH during the titration of 40 ml of 0.1000 M of the weak acid, Propanoic acid (HPr); Ka = 1.3 x 10-5 after the addition of the following volumes of 0.1000 M NaOH
0.00 mL 30.0 mL 40.0 mL 50.0 mL
a. 0.0 mL NaOH added
+
-53a
init
-[H O ][Pr ] (X)(X)K = = 1.3 10
[HPr] [HPr]
+ -2 3HPr(aq) + H O H O (aq) + Pr (Aq)
init dissoc init[HPr] = [HPr] - [HPr] [HPr]+ -
3[H O ] = [Pr ] = x
2 + 2 -5
3 a initX = [H O ] = K × HPr = 1.3 × 10 × 0.1000 M
+ -3
3[H O ] = 1.14 10 M Con’t3pH = - log[H O ] = 2.94
1/23/2015 51
Practice Problem (Con’t)b. 30.00 mL 0.1000 M NaOH added
For each mol of NaOH that reacts, 1 mol Pr- forms
Last line shows new initial amounts of HPr & Pr- that react to attain a “new” equilibrium
At equilibrium:
Original moles of HPr = 0.0400L x 0.1000 M = 0.004000 mol HPr-moles of NaOH added = 0.03000 L x 0.1000 M = 0.003000 mol OH
Amount (mol) HPr(aq) + OH-(aq) → Pr- + H2O(l)
Initial 0.004000 0.003000 0 -
Change - 0.003000 -0.003000 + 0.003000 -
Final 0.001000 0 0.003000 -
-
[HPr] 0.001000 mol = = 0.3333
[Pr ] 0.003000 mol
+3
a
-[H O ][Pr ]K =
[HPr]+ -5 -6
3 a -
[HPr][H O ] = K × = (1.3 x 10 ) (0.3333) = 4.3 x 10 M
[Pr ]
+ -6
3pH = - log[H O ] = - log(4.3 10 ) = 5.37 Con’t
1/23/2015 52
Practice Problem (Con’t)c. 40.00 mL NaOH (0.00400 mol) added
Reaction is at “equivalence point” All the HPr has reacted, leaving 0.004000 mol Pr-
Calculate concentration of Pr-
Calculate Kb
[HPr] = [OH-]
- 0.004000 mol[Pr ] = = 0.05000 M
0.04000 L + 0.004000 L-14
-10wb -5
a
K 1.0 10K = = = 7.7 10
K 1.3 10
- - 2
b - --
x x[HPr][OH ] [OH ]K = =
[Pr ] [Pr ]Pr
- -
b[OH ] = K x [Pr ]+ -
w 3K = [H O ][OH ]
-14
+ -9w3 - -10
b
K 1.0 10[H O ] = = 1.6 10 M
K [Pr ] (7.7 10 ) (0.05000)
+ -9
3pH = - log[H O ] = - log(1.6 10 ) = 8.80 Con’t
1/23/2015 53
Practice Problem (Con’t)d. 50 mL of 0.1000 M NaOH added
Beyond equivalence point, just excess OH- is being added
Calculation of pH is same as for strong acid-strong base titration
Calculate the moles of excess OH- from the initial concentration (0.1000 M) and the remaining volume (50.0 mL - 40.0 mL)
+ -w3 -
K moles of excess OH -[H O ] = where [OH ] =
[OH ] total volume
- moles of excess OH - 0.001000 mol[OH ] = = = 0.01111 M
total volume 0.04L + 0.05L
-Moles excess OH = (0.1000 M)(0.05000 L - 0.04000 L) = 0.001000 mol
-14+ -13w
3 -
K 1.0 10[H O ] = = = 9.0 10
[OH ] 0.01111
+ -13
3pH = - log[H O ] = - log(9.0 10 ) = 12.05
1/23/2015 54
Weak Base-Strong Acid Titration Neutralization reactions between weak bases and strong
acids (net ionic equation shown below)
NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l) KN=?
base acid salt water
Above equilibrium also lies very far to the right; the two equilibria below can be added together to provide the overall neutralization reaction shown above
NH3(aq) + H2O(aq) NH4+ + OH-(aq) Kb = 1.8 x 10-5
H3O+(aq) + OH-(aq) 2 H2O(l) 1/Kw = 1.0 x 1014
KN = Kb X 1/Kw = (1.8 x 10-5)(1 x 1014) = 1.8 x 109 (very large!)
Weak bases react completely with strong acids
1/23/2015 55
Weak Base-Strong Acid Titration
Curve for Titration of a Weak Base by a Strong Acid
1/23/2015 56
Polyprotic Acid Titrations Polyprotic Acids have more than one ionizable
proton Except for Sulfuric Acid (H2SO4), the common
polyprotic acids are all weak acids Successive Ka values for a polyprotic acid differ by
several orders of magnitude The first H+ is lost much more easily than
subsequent ones All “1st” protons (H+) are removed before any of the
“2nd” protons - +
2 3 2 3 3H SO (aq) + H O(l) HSO (aq) + H O (aq)
- 2- +
3 2 3 3HSO (aq) + H O(l) SO (aq) + H O (aq)
-2
a1 a1K = 1.4 x 10 and pK = 1.85
-8
a2 a2K = 6.5 x 10 and pK = 7.19
1/23/2015 57
Polyprotic Acid Titrations Each mole of H+ is titrated separately In a diprotic acid two OH- ions are required to
completely remove both H+ ions All H2SO3 molecules lose one H+ before any HSO3
- ions lose a H+ to form SO3
-
- -- 2-1 mol OH 1 mol OH
2 3 3 3H SO HSO SO
Titration curves looks like two weak acid-strong base curves joined end-to-end
HSO3- is the conjugate
base of H2SO3 (Kb = 7.1x10-13)
HSO3- is also an acid
(Ka=6.5x10-8) dissociating to form its conjugate base SO3
2-
1/23/2015 58
Slightly Soluble Compounds Most solutes, even those called “soluble,”
have a limited solubility in a particular solvent
In a saturated solution, at a particular temperature, equilibrium exists between the dissolved and undissolved solute
Slightly soluble ionic compounds, normally called “insoluble,” reach equilibrium with very little of the solute dissolved
Slightly soluble compounds can produce complex mixtures of species
Discussion here is to assume that the small amount of a slightly soluble solute that does dissolve, dissociates completely
1/23/2015 59
Equilibria - Slightly Soluble Ionic Compounds
Solubility Rules for Ionic Compounds in Water
1/23/2015 60
Equilibria - Slightly Soluble Ionic Compounds
Equilibrium exists between solid solute and the aqueous ions
PbSO4(s) Pb2+(aq) + SO42-(aq)
Set up “Reaction Quotient”
Note: concentration of a solid - [PbSO4(s)] = 1
Define Solubility Product
When solid PbSO4 reaches equilibrium with Pb2+ and SO4
2- at saturation, the numerical value of Qsp attains a constant value called the Solubility-Product constant (Ksp)
2+ 2-
4
4
c[Pb ][SO ]
Q = [PbSO (s)]
sp spQ at saturation = K = Solubility - Product constant
2+ 2-
4 4cQ PbSO (s) = Pb SO
2+ 2-
4 4sp cQ = Q [PbSO (s)] = [Pb ][SO ]
1/23/2015 61
Equilibria - Slightly Soluble Ionic Compounds
The Solubility Product Constant (Ksp)
Value of Ksp depends only on temperature, not individual ion concentrations
Saturated solution of a slightly soluble ionic compound, MpXq, composed of ions Mn+ and Xz-, the equilibrium condition is:
n+ p z- q
sp spQ = [M ] [X ] = K
1/23/2015 62
Equilibria - Slightly Soluble Ionic Compounds
Single-Step Process
Multi-Step Process
2+ -
2Cu(OH) (s) Cu (aq) + 2OH (aq)
2+ - 2
spK = [Cu ][OH ]
2+ 2-MnS(s) Mn (aq) + S (aq)
2- - -
2S (aq) + H O(l) HS (aq) + OH (aq)
2+ - -
2MnS(s) + H O(l) Mn (aq) + HS (aq) + OH (aq)
2+ - -
spK = [Mn ][HS ][OH ]
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Equilibria - Slightly Soluble Ionic Compounds
1/23/2015 64
Practice ProblemDetermine Ksp from solubility
The solubility of PbSO4 at 25oC is 4.25 x10-3 g /100 mL solution. Calculate the Ksp
Convert solubility to molar solubility
Some PbSO4 dissolves into Pb2+ & SO42-
2+ 2-
4 4PbSO (s) = Pb (aq) + SO (aq)
4 4
4
0.00425 g PbSO 1 mol PbSO1000 mlMolar Solubility = x x
100 mL 1L 303.3 g PbSO
2+ 2- -4 2 -8
sp 4K = [Pb ][SO ] = (1.40 10 ) = 1.96 10
2+ 2-
4[Pb ] = SO
2+ 2- -4
4[Pb ] = SO = 1.40 10 M
-4
4= 1.40 10 M PbSO
1/23/2015 65
Practice ProblemDetermine Solubility from Ksp
Determine the solubility of Calcium Hydroxide, Ca(OH)2
Ksp for Ca(OH)2 = 6.5 x 10-6
Set up reaction table (S = molar solubility)
2+ -
2Ca(OH) (s) Ca (aq) + 2OH (aq)
Concentration (M) Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq)
Initial 0 0
Change +S +2S
Equilibrium S 2S2+ - 2 2 2 3 -6
spK = [Ca ][OH ] = (S)(2S) = (S)(4S ) = 4S = 6.5 x 10
-62+ -23
6.5x10[Ca ] = S = = 1.2 x 10 M = 0.012 M
4
- -2 -2[OH ] = 2S = 2 1.2 10 = 4.2 10 M = 0.042 M
1/23/2015 66
Equilibria - Slightly Soluble Ionic Compounds
The Effect of the Common Ion
The presence of a “Common Ion” decreases the solubility of a slightly soluble ionic compound
Add some Na2CrO4, a soluble salt, to the saturated PbCrO4 solution
Concentration of CrO42- increases
Some of excess CrO42- combines with Pb2+ to
form PbCrO4(s)
Equilibrium shifts to the “Left” This shift “reduces” the solubility of PbCrO4
2+ 2-
4 4PbCrO (s) Pb (aq) + CrO (aq)€
2+ 2- -13
sp 4K = [Pb ][CrO ] = 2.3 x 10
1/23/2015 67
Practice ProblemWhat is the solubility of Ca(OH)2 in 0.10 M Ca(NO3)2? Ksp Ca(OH)2 = 6.5 x 10-6
The addition of the common ion, Ca2+, should lower the solubility of Ca(OH)2 Note: (Compare to result from slide 65)
Since Ksp is small, S << 0.10 then, 0.10 M + S 0.1 M
2+ -
2Ca(OH) (s) Ca (aq) + 2OH (aq)
Concentration (M) Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq)
Initial 0.10 0
Change +S +2S
Equilibrium 0.10 + S = 0.1 2S
2+ - 2 -6 2
spK = [Ca ][OH ] = 6.5 x 10 (0.1)(2S)
-62 -56.5 10
4S = = 6.5 100.10
-52+ -36.5 x 10
S = [Ca ] = = 4.03 10 M4
-34.0 x 10 M 100 = 4.0% < 5%
0.10 M Assumption : S << 0.10 is valid
2+ 2+
this prob prev prob[Ca ] vs. [Ca ] ( 0.00403 M vs. 0.012 M)
1/23/2015 68
Equilibria - Slightly Soluble Ionic Compounds
The Effect of pH on Solubility The Hydronium Ion (H3O+) of a strong acid
increases the solubility of a solution containing the anion of a weak acid (HA-)
Adding some strong acid to a saturated solution of Calcium Carbonate (CaCO3) introduces large amount of H3O+ ion, which reacts immediately with the CaCO3 to form the weak acid HCO3
-
Additional H3O+ reacts with the HCO3- to form
carbonic acid, H2CO3, which immediately decomposes to H2O and CO2
2+ 2-
3 3CaCO (s) Ca (aq) + CO (aq)€2- + -
3 3 3 2CO (aq) + H O HCO (aq) + H O(l)€
- +
3 3 2 3 2 2 2HCO (aq) + H O H CO (aq) + H O(l) CO (g) + 2H O(l)
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Equilibria - Slightly Soluble Ionic Compounds
The equilibrium shifts to the “Right” and more CaCO3 dissolves – increased solubility
The overall reaction is:
Adding H3O+ to a saturated solution of a compound with a strong acid anion – AgCl
Chloride ion, Cl-, is the conjugate base of a strong acid (HCl)
It coexists with water, i.e. does not react with water
There is “No” effect on the equilibrium
3 3+ +
2+ 2- -
3 3 3
2+
2 3 2 2
H O H OCaCO (s) Ca + CO HCO
H CO CO (g) + H O + Ca
€
1/23/2015 70
Practice ProblemWrite balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of the following ionic compounds
Bromide ion does not react with water
Lead Bromide (PbBr2)
2+ -
2PbBr Pb + 2Br€
Addition of an Acid has No effect
-Bromide (Br ) is the anion of the strong acid HBr
1/23/2015 71
Practice Problem Cu(II) Hydroxide (Cu(OH)2
-
2more Cu(OH) dissolves to replace OH
2+ -
2Cu(OH) (s) Cu (aq) + 2OH (aq)€
Increases Solubility
-
2OH is anion of H O, a very weak acid
+
3 2It reacts with added H O to form water (H O)
- +
3 2OH (aq) + H O (aq) 2H O
1/23/2015 72
Practice ProblemIron(II) Sulfide (FeS)
2+ - -
2FeS(s) + H O(l) Fe (aq) + HS (aq) + OH (aq)€
+ - -
3Additional H O reacts with both HS and OH (both weak acid anions)
- +
3 2 2HS (aq) + H O (aq) H S(aq) + H O(l)
- +
3 2OH (aq) + H O 2H O(l)
+
3Additional H O Increases Solubility
) - -
2Some of the FeS dissolves in water (H O to form HS and OH
- -More FeS dissolves to replace the HS & OH
1/23/2015 73
Practice Problem Calcium Fluoride (CaF2)
2+ -
2CaF (s) Ca (aq) + 2F (aq)€
-F is the conjugate base of a weak acid (HF)
- +
3F reacts with added H O to form the acid and water
2The solubility of CaF increases
- +
3 2F (aq) + H O (aq) HF(aq) + H O(l)
2-More CaF dissolves to replace the F
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Equilibria - Slightly Soluble Ionic Compounds
Predicting Formation of a Precipitate
Qsp = Ksp when solution is “saturated”
Qsp > Ksp solution momentarily “supersaturated”
Additional solid precipitates until Qsp = Ksp again
Qsp < Ksp solution is “unsaturated”
No precipitate forms at that temperature
More solid dissolves until Qsp = Ksp
1/23/2015 75
Practice ProblemDoes CaF2 precipitate when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?
Ksp (CaF2) = 3.2 x 10-11
Ions present: Ca2+ Na+ NO3- F-
Note: All sodium & nitrate salts are soluble2+ 2+ 2+moles Ca = 0.30 M Ca x 0.100L = 0.030 mol Ca
2+2+ 2+
init
0.030 mol Ca[Ca ] = = 0.10 M Ca
0.100 L + 0.200 L moles F = 0.060 M F x 0.200L = 0.012 mol F
-- -
init
0.012 mol F[F ] = = 0.040 M F
0.100 L + 0.200 L2+ - 2 2 -4
sp init initQ = [Ca ] [F ] = (0.10)(0.040) = 1.6 10-4 -11
sp spQ > K (1.6 10 > 3.2 10 )
2 sp sp CaF will precipitate until Q = K
1/23/2015 76
Equilibria - Slightly Soluble Ionic Compounds
Selective Precipitation of Ions Separation of one ion in a solution from another Exploit differences in the solubility of their
compounds The Ksp of the less soluble compound is much smaller
than the Ksp of the more soluble compound
Add solution of precipitating ion until the Qsp value of the more soluble compound is almost equal to its Ksp value
The less soluble compound continues to precipitate while the more soluble compound remains dissolved, i.e., the Ksp of the less soluble compound is always being exceeded, i.e. precipitation is occurring
At equilibrium, most of the ions of the less soluble compound have been removed as the precipitate
1/23/2015 77
Practice ProblemA solution consists of 0.20 M MgCl2 and 0.10 M CuCl2
Calculate the [OH-] that would separate the metal ions as their hydroxides
Ksp of Mg(OH)2 = 6.3 x 10-10 Ksp of Cu(OH)2 = 2.2 x 10-
20
Mg(OH)2 is 1010 more soluble than Cu(OH)2; thus Cu(OH)2 precipitates first
Solve for [OH-] that will just give a saturated solution of Mg(OH)2
This concentration of [OH-] is more than enough for the less soluble CuCl2 to reach its Ksp and precipitate completely.
Con’t
1/23/2015 78
Practice ProblemWrite the equations and ion-product expressions
Calculate [OH-] of saturated Mg(OH)2 solution
This is the maximum amount of [OH-] that will not precipitate Mg2+ ion
Calculate [Cu2+] remaining in solution
Since initial [Cu2+] was 0.10M, virtually all Cu2+ is precipitated
2+ - 2+ - 2
2 spMg(OH) (s) Mg (aq) + 2OH (aq) K = [Mg ][OH ]€2+ - 2+ - 2
2 spCu(OH) (s) Cu (aq) + 2OH (aq) K = [Cu ][OH ]€
-10sp- -5
2+
K 6.3 10[OH ] = = = 5.6 10 M
[Mg ] 0.20
-20sp2+ -12
- 2 -5 2
K 2.2 x 10[Cu ] = = = 7.0 x 10 M
[OH ] (5.6 x 10 )
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Equilibria - Slightly Soluble Ionic Compounds
Equilibria Involving Complex Ions The product of any Lewis acid-base reaction is called
an “Adduct”, a single species that contains a new covalent bond
A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands
Ionic ligands – OH- Cl- CN-
Molecular ligands – H2O CO NH3
Ex Cr(NH3)63+
Cr3+ is the central metal ion
NH3 molecules are molecular ligands Metal acts as Lewis Acid by accepting electron pair;
ligands acts as Lewis base by donating electron pair All complex ions are Lewis adducts
A + : B A - B (Adduct)
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Equilibria - Slightly Soluble Ionic Compounds
Complex Ions Acidic Hydrated metal ions are complex ions with
water molecules as ligands When the hydrated cation is treated with a
solution of another ligand, the bound water molecules exchange for the other ligand
At equilibrium
Water is constant in aqueous reactions Incorporate in Kc to define Kf (formation constant)
2+ 2+
2 4 3 3 4 2M(H O) (aq) + 4NH (aq) M(NH ) (aq) + 4H O(l)
2+ 4
3 4 2c 2+ 4
2 4 3
[M(NH ) ][H O]K =
[M(H O) ][NH ]
2+
c 3 4f 4 2+ 4
2 2 4 3
K [M(NH ) ]K = =
[H O] [M(H O) ][NH ]
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Equilibria - Slightly Soluble Ionic Compounds
Complex Ions (Con’t) The actual process is stepwise, with ammonia
molecules replacing water molecules one at a time to give a series of intermediate species, each with its own formation constant (Kf)
The sum of the 4 equations gives the overall equation
The product of the individual formation constants gives the overall formation constant
The Kf for each step is much larger than “1” because “ammonia” is a stronger Lewis base than H2O
Adding excess NH3 replaces all H2O and all M2+ exists as M(NH3)4
2+
2+ 2+
2 4 3 2 3 3 2M(H O) (aq) + NH (aq) M(H O) (NH ) (aq) + H O(l)2+
2 3 3f1 2+
2 4 3
[M(H O) (NH )K =
[M(H O) ][NH ]
f f1 f2 f3 f4K = K x K x K x K
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Practice ProblemWhat is the final [Zn(H2O)4
2+] after mixing 50.0 L of0.002 M Zn(H2O)4
2+ and 25 L of 0.15 M NH3
Determine Limiting Reagent
2+ 2+
2 4 3 3 4 2Zn(H O) + 4NH (aq) Zn(NH ) + 4H O(l)
Con’t
2+ 8f 3 4K of Zn(NH ) = 7.8 10
2
2
3 4
f 4
2 4 3
[Zn(NH )K =
[Zn(H O) ][NH ]
3Moles NH = 25 L×0.15 mol / L = 3.75 mol
2+2 4thus, Zn(H O) is limiting
2+2 4Moles Zn(H O) = 50.0 L×0.002 mol / L = 0.100 mol
2+3 2 4Moles NH (3.75) >> 4 Moles Zn(H O) (4 0.1)
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Practice Problem (Con’t) Find the initial reactant concentrations:
2+ -3
2 4 init
50.0 L 0.0020 M[Zn(H O) ] = = 1.3 10 M
50.0 L + 25.0 L
-2
3 init
25.0 L 0.15 M[NH ] = = 5.0 10 M
50.0 L + 25.0 L
Con’t
1/23/2015 84
Practice Problem (Con’t)Assume most Zn(H2O)4
2+ is converted to Zn(NH3)42+
Solve for x, the [Zn(H2O]42+ remaining at equilibrium
Concentration (M) Zn(H2O)42+(aq) + 4NH3(aq) ⇄ Zn(NH3)4
2+(aq) + 4H2O(l)
Initial 1.3 x 10-3 5.0 x 10-2 0
Change (-1.3 x 10-3)4 x -1.3 x 10-3
(-5.2 x 10-3)(+1.3 x 10-3)
Equilibrium X 4.5 x 10-2 1.3 x 10-3
2+ -383 4
f 2+ 4 -2 4
2 4 3
[Zn(NH ) ] 1.3 10K = = 7.8 10
[Zn(H O) ][NH ] (x) (4.5 10 )
2+ -7
2 4x = [Zn(H O) ] 4.1 10 M
Kf is very large; [Zn(H2O)42+ should be “low”
-3 -33 reacted[NH ] 4(1.3 10 M) 5.2 10 M
2+ -33 4[Zn(NH ) ] 1.3 10 M
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Complex Ions – Solubility of Precipitates
Increasing [H3O+] increases solubility of slightly soluble ionic compounds if the anion of the compound is that of a weak acid
A Ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation
When 1.0 M NaCN is added to the above solution, the CN- ions act as ligands and react with the small amount of Zn2+(aq) to form a complex ion
2+ - -
2ZnS(s) + H O(l) Zn (aq) + HS (aq) + OH (aq)
2+ - 2-
4Zn (aq) + 4CN (aq) Zn(CN) (aq)
-22
spK = 2.0 10
19
fK = 4.2 10
1/23/2015 86
Complex Ions – Solubility of Precipitates
Add the two reactions and compute the overall K2+ - -
2ZnS(s) + H O(l) Zn (aq) + HS (aq) + OH (aq)€2+ - 2-
4Zn (aq) + 4CN (aq) Zn(CN) (aq)€
- 2- - -
2 4ZnS(s) + 4CN (aq) + H O(l) Zn(CN) + HS (aq) + OH (aq)
-22 19 -3
overall sp fK = K K = (2.0 10 ) (4.2 10 ) = 8.4 10
The overall equilibrium constant, Koverall , is more than a factor of 1019 larger than the original Ksp (2.0 x 10-22)
This reflects the increased amount of ZnS in solution as Zn(CN4
2-)
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Practice Problem
Calculate the solubility of Silver Bromide (AgBr) in water (H2O) and 1.0 M of photographic “Hypo” – Sodium Thiosulfate (Na2S2O3), which forms the “complex” ion Ag(S2O3)2
3-
a. Solubility AgBr in Water
Con’t
-13 3- 13
sp f 2 3 2K (AgBr) = 5 10 K (Ag(S O ) ) = 4.7 10
+ -AgBr(s) Ag (aq) + Br (aq)
-13 + -
spK = 5.0 10 = [Ag ][Br ]
+ -
dissolvedS = [AgBr] = [Ag ] = [Br ]
-13 +
spK = 5.0 10 = [Ag ][Br-] = S x S
-7S = 7.1 10 M
-13
spK = 5 10
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Practice Problem (con’t)Solubility AgBr in 1.0 M Hypo - Sodium Thiosulfate (Na2S2O3)
Write the overall equation:
Calculate Koverall (product of Ksp & Kf):
+ -AgBr(s) Ag (aq) + Br (aq)€+ 2- 3-
2 3 2 3 2Ag (aq) + 2S O (aq) Ag(S O ) (aq)€2- 3- -
2 3 2 3 2AgBr(s) + 2S O (aq) Ag(S O ) (aq) + Br (aq)
]
3- --13 132 3 2
overall sp f2-
2 3
[Ag(S O ) ][Br ]K = = K K = (5.0 × 10 )(4.7 × 10 ) = 24
[(S O )
1/23/2015 89
Practice Problem (Con’t) Set Up the Reaction Table
3-
dissolved 2 3 2S = [AgBr] = [Ag(S O ) ]
Concentration (M) AgBr(s) + 2S2O32-(aq) ⇄ Ag(S2O3)2
3-(aq) + Br-(l)
Initial 1.0 0 0
Change -2S +S +S
Equilibrium 1.0 - 2S S S
243- 2
2 3 2overall 2- 2
2 3
[Ag(S O ) ][Br-] SK = =
[(S O ) ] (1.0 M - 2S)
S = 24 = 4.9
1.0 - 2SS = 4.9 - 2S(4.9) S + 2S(4.9) = 4.9
3-
2 3 2S = Solubility Ag(S O ) or AgBr = 0.45 M
S(1 + 2× 4.9) = 4.94.9
S = = 0.45 M1 + 9.8)
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Ionic Equilibria in Aqueous Solutions
Equilibria Involving Complex Ions Amphoteric Oxides & Hydroxides (Recall Chapter 8)
Some metals and many metalloids form oxides or hydroxides that are amphoteric; they can act as acids or bases in water
These compounds generally have very little solubility in water, but they do dissolve more readily in acids or bases Ex. Aluminum Hydroxide
Al(OH)3(s) ⇆ Al3+(aq) + 3OH-(ag)
Ksp = 3 x 10-34 (very insoluble in water)
In acid solution, the OH- reacts with H3O+ to form water
3H3O+(ag) + 3OH-(aq) 6H2O(l)
Al(OH)3(s) + H3O+(aq) Al3+(aq) + 6H2O(l)
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Ionic Equilibria in Aqueous Solutions
Equilibria Involving Complex Ions
Aluminum Hydroxide in basic solution
Al(OH)3(s) + OH-(aq) Al(OH)4-(aq)
The above reaction is actually a much more complex situation, involving multiple species
When dissolving an aluminum salt, such as Al(NO3), in a strong base (NaOH), a precipitate forms initially and then dissolves as more base is added
The formula for hydrated Al3+ is Al(H2O)63+
Al(H2O)63+ acts as a “weak polyprotic acid and
reacts with added OH- in a stepwise removal of the H2O ligands attached to the hydrated Al
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Ionic Equilibria in Aqueous Solutions
Amphoteric Aluminum Hydroxide in basic solution
Al(H2O)63+(aq) + OH-(aq) ⇆ Al(H2O)5OH2+(aq) + H2O(l)
Al(H2O)52+(aq) + OH-(aq) ⇆ Al(H2O)4(OH)2
+(aq) + H2O(l)
Al(H2O)4+(aq) + OH-(aq) ⇆ Al(H2O)3(OH)3(s) + H2O(l)
Al(H2O)3(OH)3(s) is more simply written Al(OH)3(s)
As more base is added, a 4th H+ is removed from a H2O ligand and the soluble ion Al(H2O)2(OH)4
-(aq) forms
Al(H2O)3(OH)3(s) + OH-(aq) ⇆ Al(H2O)2(OH)4-(aq)
The ion is normally written as Al(OH)4-(aq)
1/23/2015 93
Equation Summary- +
3[HA ][H O ]Ka =
[HA]
Concentration- A(aq) + B(aq) A-(aq) + H2O(l)
Initial Ci - - - 0
- - +X +X
Equilibrium Cf - - X X
+ -2 3HA + H O H O + A€
+ -3
a
[H O ][A ]K =
[HA]
a[base]
pH = pK + log[acid]
-
a a a a
[A ]pH = pK + log = pK + log1 = pK + 0 = pK
[HA]
-[A ]When : = 1
[HA]
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Equation Summary
n+ p z- q
sp spQ = [M ] [X ] = K
-14
w a bK = K K = 1.0 10
wpK = pH + pOH = 14
+ -
b[BH ][OH ]
K = [B]
2142
nw3 3
- -+ +[H O] 1 1
K = = = = 1×10K[H O ][OH ] [H O ][OH ]
2 3
+ -2H O H O + OH
3 2
+ -H O + OH 2H O
-143w 32
2
+ -+ -[H O ][OH ]
K = = [H O ][OH ] = 1×10 [H O]