Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay,

AqueousEquilibria

Chapter 17Additional Aspects of Aqueous Equilibria

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

John D. BookstaverSt. Charles Community CollegeSt. Peters, MO2006, Prentice Hall, Inc.Modified by S.A. Green, 2006

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Control of solution composition

Chap 16: What is the pH after adding X to water?

Chap 17:How can we control the pH of a solution?How can we determine acid/base concentration?

Chap 4:Solubility of ionic compounds.

Chap 17:How soluble are they?

Buffers

Titrations

K sp=solubility product

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The Common-Ion Effect

• Consider a solution of acetic acid:

• If acetate ion (CH3O2-) is added to the

solution, Le Châtelier says the equilibrium will shift to the left <––.

Where can we get CH3O2- ions?

H3C

C

O

OH

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The salt completely dissociates.

So, adding the salt = adding CH3O2- (acetate) ions.

What happens to the pH?


Where can we get CH3O2- ions?

Add them in the form of a salt:

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Calculate the pH of a 1.0 L solution containing 0.30 moles of acetic acid, C2H3O2H, and 0.30 moles of sodium acetate C2H3O2Na, at 25°C.

Ka for acetic acid at 25°C is 1.8 10-5.


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Use the same method as in Chapter 16:ICE + equilibrium expression

But, adjust initial concentrations to account for the salt.


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Use the ICE table:

[C2H3O2], M [H3O+], M [C2H3O2−], M

Initial 0.30 0 0.30

Change –x +x +x

Equilibrium 0.30 – x x 0.30+x


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Suppose x is small relative to 0.30:

Ka= x1.8 10-5 = x = [H+]pH=–log(1.8 10-5)=4.74

Check: is approximation ok?


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Note:

When [HA] = [A-] then: [H3O+] = Ka

The Common-Ion Effect: Buffering

When [H3O+] is small compared to [HA] & [A-]

then [H3O+] = Ka times the ratio [HA]/[A-]

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Buffers:

• Solutions of a weak conjugate acid-base pair.

• They are particularly resistant to pH changes, even when strong acid or base is added.

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Buffers

If a small amount of hydroxide is added to an equimolar solution of HF and NaF, for example, the HF reacts with the OH– to make F– and water.

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Buffers

If acid is added, the F– reacts to form HF and water.

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Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

[H3O+] [A−][HA]

Ka =

HA + H2O H3O+ + A−

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Buffer Calculations

Rearranging slightly, this becomes

[A−][HA]

Ka = [H3O+]

Taking the negative log of both side, we get

[A−][HA]

–log Ka = –log [H3O+] + −log

pKa

pHacid

base

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Buffer Calculations

• SopKa = pH –log

[base][acid]

• Rearranging, this becomes

pH = pKa + log[base][acid]

• This is the Henderson–Hasselbalch equation.

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Henderson–Hasselbalch Equation

What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate?

Ka for lactic acid is1.4 10-4.

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Henderson–Hasselbalch Equation

pH = pKa + log[base][acid]

pH = –log (1.4 10-4) + log(0.10)(0.12)

pH = 3.85 + (–0.08)

pH = 3.77

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pH Range

• The pH range is the range of pH values over which a buffer system works effectively.

• It is best to choose an acid with a pKa close to the desired pH.

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When Strong Acids or Bases Are Added to a Buffer…

…it is safe to assume that all of the strong acid or base is consumed in the reaction.

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Addition of Strong Acid or Base to a Buffer

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

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Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

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Before the reaction, since

mol HC2H3O2 = mol C2H3O2–

pH = pKa = –log (1.8 10-5) = 4.74

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Calculating pH Changes in BuffersThe 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH-(aq) C2H3O2-(aq) + H2O(l)

HC2H3O2 C2H3O2− OH−

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction 0.280 mol 0.320 mol ≈0.000 mol

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Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log(0.320)(0. 200)

pH = 4.74 + 0.06

pH = 4.80

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Titration

A known concentration of base (or acid) is slowly added to a solution of acid (or base).

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TitrationA pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Equiv point: moles acid=moles base

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Titrations

Types: 1. “strong-strong”Strong acid + strong baseStrong base + strong acid

2. “weak-strong”Weak acid + strong baseWeak base + strong acid

Why titrate?1. How much acid or

base is present .

2. What is pKa (pKb) of weak acid (base).(we will not do this)

Usual question asked: What is the pH after X mL of Y has been added?

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Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.

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Just before and after the equivalence point, the pH increases rapidly.

--> transition from acidic to basic state

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At the equivalence point:

moles acid = moles base. pH=7The solution contains

only water and the salt from the cation of the base and the anion of the acid.

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As more base is added, the increase in pH again levels off.

Movie….

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Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH.

• The pH at the equivalence point is >7.

Phenolphthalein is commonly used as an indicator in these titrations.

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Before the equivalence point:

the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present.

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With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

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Titration of a Weak Base with a Strong Acid

• The pH at the equivalence point in these titrations is < 7.

• Methyl red is the indicator of choice.

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SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 10–5).Solution

Write reaction Ask: Is this strong-strong or weak-strong titration?

Are we before, at, or after equivalence point?

Need to find:# moles of weak acid present initially# moles of base addedConcentrations of acid and conjugate base

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SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 10–5).Need to find:# moles of weak acid present initially# moles of base addedConcentrations of acid and conjugate base

The 4.50 10–3 mol of NaOH consumes 4.50 10–3 mol of HC2H3O2:

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SAMPLE EXERCISE 17.7 continued

The total volume of the solution is

The resulting molarities of HC2H3O2 and C2H3O2– after the reaction are therefore

Equilibrium Calculation: The equilibrium between HC2H3O2 and C2H3O2– must obey the equilibrium-constant

expression for HC2H3O2:

Solving for [H+] gives

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 10–5).

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Titrations of Polyprotic Acids

In these cases there is an equivalence point for each dissociation.

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Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42-(aq)

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Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42-]

where the equilibrium constant, Ksp, is called the solubility product.

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Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

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Factors Affecting Solubility

• The Common-Ion Effect If one of the ions in a solution equilibrium

is already dissolved in the solution, the equilibrium shifts to the left and the solubility of the salt decreases.

BaSO4(s) Ba2+(aq) + SO42-(aq)

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• pH If a substance has a

basic anion, it is more soluble in an acidic solution.

Substances with acidic cations are more soluble in basic solutions.

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Factors Affecting Solubility• Complex Ions

Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

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• Complex IonsThe formation

of these complex ions increases the solubility of these salts.

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• Amphoterism Amphoteric metal

oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

Examples of such cations are Al3+, Zn2+, and Sn2+.

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Will a Precipitate Form?

• In a solution, If Q = Ksp, the system is at equilibrium

and the solution is saturated. If Q < Ksp, more solid will dissolve until

Q = Ksp.

If Q > Ksp, the salt will precipitate until Q = Ksp.

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Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

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Chapter 17 problems

17.22 A buffer is prepared by adding 5.0 g of NH3 and 20.0 g NH4

+ to enough water to make 2.50 L of solution.

(a) pH=?(b) (b) Write complete ionic equation for the reaction

that occurs when a small amount of concentrated nitric acid is added.

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Chapter 17 problems

17.23 How would you make about 1 L of a buffer with pH=3.50? The following 0.10 M solutions are available:

HCO2

HC2H3O2

H3PO4

NaCHO2

NaC2H3O2

NaH2PO4

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Chapter 17 problems

17.41 Consider the titration of 35.0 mL of 0.15 M acetic acid with 0.15 M NaOH. What is the pH after the following amounts of titrant have been added?

34.5 mL35.0 mL35.5 mL

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Chapter 17 problems

17.48 The molar solubility of PbBr2 is 1.0 x 10-2 mol/L.

Cacluate Ksp.

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Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay,