IB Chemistry ABS — Introductionsemmchem.cmswiki.wikispaces.net/file/view/ibchabs.pdf/482713820/... · IB Chemistry — Acids Bases and Salts 1 IB Chemistry ABS — Introduction

IB Chemistry — Acids Bases and Salts

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IB Chemistry ABS — Introduction An acid was initially considered a substance that would produce H+ ions in water. The Brønsted-Lowry definition of an acid is a species that can donate an H+ ion to any other species (not just water). The Lewis definition defines acids by the electron pairs they accept in a reaction. Most acids have a weakly bonded H atom. A base was initially defined as a species that produces OH– ions in water. Alkalis refer to bases that dissolve in water to produce some OH– ions in water. The Brønsted-Lowry definition describes a base as a proton acceptor. The Lewis definition describes a base as an electron pair donor. Many common bases have an OH– anion.

Examples of Acid/Base reactions: 1) Traditional acid-base reaction HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) CH3COOH(aq) + KOH(aq) H2O(l) + KCH3COO(aq) Neutralization reaction produces water and a salt. 2) Brønsted-Lowry reaction HBr(aq) + NH3(aq) NH4Br(aq) HBr is the proton donator, and NH3 is the proton acceptor (does not have OH– as an anion). 3) An acid will react with some metals to form Hydrogen gas (wait until redox to see which metals): 2 HCl(aq) + Mg(s) H2(g) + MgCl2(aq) This can be used as a test for an acid.


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4) An acid that is reacted with a carbonate or a hydrogen carbonate is a traditional acid/base reaction that produces carbon dioxide gas: 2 HNO3(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca(NO3)2(aq) HI(aq) + NaHCO3(s) H2O(l) + CO2(g) + NaI(aq) This is used as a test for carbonates in minerals such as limestone (CaCO3) 5) Lewis Acid/Base reaction The lewis acid, BF3 (electron poor, not a full octet) is accepting the electron pair from the lewis base F– (electron rich) that is donating the electron pair. A lewis acid-base reaction usually produces a covalent bond where both of the electrons in the bond came from one species. Complexes with ligands are usually lewis acid-base reactions. Brønsted-Lowry acids and bases must come in pairs. Many of them are also reactions that from an equilibrium. HF + H2O F– + H3O+

Here HF is acting as the acid and water is acting as the base. However, this is a reversible reaction. HF and F– are called a conjugate acid/base pair. Similarly, H2O and H3O+ are also a conjugate acid/base pair. A conjugate acid/base pair differ by one proton. An acid as a reactant, must have a base as a product.


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Ex 1 Identify the acid-base conjugate pairs in the following equilibria: PO4

3- + H2O HPO42- + OH–

NH4

+ + H2O H3O+ + NH3 HCN + H2O H3O+ + CN– Follow Up Problems 18.4, 13 Problems 18.2, 5, 7, 37, 38, 43, 45, 47, 49, 133, 137

IB Chemistry ABS — Water and pH Most of our acid/base chemistry is carried out in aqueous environments. Water is an amphiprotic substance. An amphiprotic substance can act as an acid and a base. In previous examples, we saw water acting as a Brønsted-Lowry acid and a base: PO4

3- + H2O HPO42- + OH–

NH4

+ + H2O H3O+ + NH3 Water can act as an acid and a base with itself: auto-ionization. 2 H2O(l) H3O+

(aq) + OH–(aq)

Every water solution will have some H3O+ and some OH– The equilibrium constant for this reaction is: Kw = [H3O+][OH–] At 25°C, Kw = 1.0 x 10-14 A pure water solution at 25°C will have Kw = [H3O+][OH–] = 1.0 x 10-14


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In pure water the only source of these ions is: 2 H2O(l) H3O+

(aq) + OH–(aq)

Then, [H3O+] = [OH–] = 1.0 x 10-7 M The solution is still neutral because: [H3O+] = [OH–] An Acid solution has [H3O+] > [OH–] A basic solution has [H3O+] < [OH–] Ex 1 Use Le Châtelier's Principal and the Kw expression to explain the changes to [H3O+] and [OH–] as an acid and a base is added to pure water. Acidity and basicity is measured by a logarithmic scale. The pH scale is based on the “power” of Hydrogen. pH = -log[H3O+] pH has no units. The pH of a neutral water solution at 25°C is -log(1.0 x 10-7M) = 7.00 Note: In pH, only numbers after the decimal are significant.


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[H3O+]

pH

Acidic or basic

1.0 M (100 M)

10–2 M

10–4 M

1.0 x 10–5 M

5.0 x 10–7 M

1.0 x 10–7 M

10–9 M

2.0 x 10–11 M

4.0 x 10–14 M

Each change of a pH value is a 10x change in acid strength.


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To convert pH back to [H3O+], [H3O+] = 10-pH

The prefix “p” means to take the negative log of a value. pKw = -logKw. pOH = -log[OH–] Kw = [H3O+][OH–] = 1.0 x 10-14 Taking the negative log of this gives: pKw = pH + pOH = 14.00

The value of Kw varies with temperature. The auto-ionization of water is an endothermic process. The equilibrium shifts to the products side as temperature increases. Ex 2 a) What is the pH of neutral water at 50°C b) The pOH of a neutral water sample is 6.40. What is the temperature of this sample? Follow Up Problems 18.2, 3, Problems 18.21, 27, 29, 36, 40

[H3O+ ]

Kw

=[H3O+ ][OH– ]

! "##### [OH– ]

pH= –log[H3O

+]! ! pOH= –log[OH

–]

pH pH + pOH = 14

! "#### pOH

[H3O+] [OH

–] pH pOH A/B

3.5 x 10–5

11.05

6.805 x 10–4

8.50

A

B


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IB Chemistry ABS — Acid Solutions Acids are characterized by the degree that they dissociate (ionize) in water. A strong acid will dissociate completely in water. It will be a strong electrolyte. A weak acid will only dissociate partially in water; it will form an equilibrium. It will be a weak electrolyte. Be careful with these words. Concentrated and dilute refer to molarity. The strong acids you need to know are: HCl, HBr, HI, H2SO4, HNO3. These will dissociate completely in water HCl(aq) + H2O(l) H3O+

(aq) + Cl–(aq)

Ex 1: Calculate the pH of a 0.200 M HCl solution. Calculate the pOH of a 7.15 x 10-5M HNO3 solution.


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A weak acid solution will form an equilibrium with water. There are many weak acids, some are given on Table 16, and in the appendix of your text. We will use a generic weak acid HA HA(aq) + H2O(l) H3O+

(aq) + A–(aq)

Ka is the equilibrium constant for a weak acid reacting with water. You only need to dissociate one proton in any question. In IB, Ka values will have units of kmol/m3. pKa values are used in the IB Data Booklet. pKa = -logKa. (no units) Often, units are omitted from Ka values. Ex 2: Find the pH of a 0.100 M HF solution. Ka = 6.8 x 10-4

Note the assumption.

Ka=

[H3O

+][A

-]

[HA]


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Ex 3: A 0.035 M solution of an unknown acid has a pH of 5.80. What is the pKa of this acid? Ex 4: How could you determine if a 0.50 M acid solution is a strong or weak acid solution?


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Ex 5: A solution of benzoic acid has a pH of 5.10. What is the concentration of the solution? Follow Up Problems 18.6, 7 Problems 18.25(a), 59, 60, 63, 65, 67, 71

IB Chemistry ABS — Basic Solutions Bases also fall into the categories of strong and weak. Strong bases are group 1 hydroxides and Ba(OH)2. There are many weak bases including ammonia and methyl amine. (Nitrogen with a lone pair of electrons) Strong bases dissociate completely in water. Ex 1: Find the pH of a 2.0 M NaOH solution. Find the pOH of a 0.015 M Ba(OH)2 solution.


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A weak base solution dissociates partially in water. NH3(aq) + H2O(l) OH–

(aq) + NH4+

(aq)

This is the equilibrium expression for a weak base reacting with water. Ex 2: A 0.25 M solution of an unknown base has a pH of 11.80. What is the dissociation constant value for this weak base? For a weak acid HA: For its conjugate base A–:

Kb=

[NH4

+][OH

-]

[NH3]

Ka=

[H3O

+][A

-]

[HA]

Kb=

[HA][OH-]

[A–]

Ka! K

b=

[H3O

+][A

-]

[HA]

[HA][OH-]

[A–]

= [H3O

+][OH

-]=K

w

K

a! K

b=K

w


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It is very important to remember to use the Ka value of the conjugate acid of the weak base. To find the Kb of HCO3

–, we must use the the Ka value of the conjugate acid: H2CO3. Ex 3: What is the pH of a 0.038 M solution of acetate ions (CH3COO–). What are the trends in acid base strengths? Follow Up Problems 18.9, 10 Problems 18.25(b), 83, 86, 88, 90, 94, 96, 98, 100, 102


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IB Chemistry ABS — Salt Solutions Many aqueous ions form acidic or basic solutions. Cations can form neutral or acidic solutions. Ammonium (NH4

+) is a weak acid, it will form an acidic solution. Cations that are small and highly charged can form acidic solutions. The water molecules have electrons pulled away from the O—H bond, making it weaker and the solution acidic. The higher the charge density, the greater the acidity. Al3+, Fe3+, and Cr3+ have notable acidity. These are small highly charged ions. Fe2+ or Cu2+ would have a much lower acidity. Alkali metals do not form acidic solutions. Cations with a 2+ charge only form very slightly acidic solutions if the ion is very small. Anions can form acidic or basic or neutral solutions. The conjugate base of a strong acid will form a neutral solution: Cl–, Br–, I–, NO3

– The anion can be the conjugate base of a weak acid: a weak base. These anions are weak bases and will make a solution basic: F–, NO2

–, CH3COO–, CO32–.

Ex 1: Indicate if the following solutions will be acidic, basic or neutral: a) NaF b) AlCl3 c) NH4NO3 d) KCH3COO e) MgI


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An anion can also be amphiprotic: HCO3–, HSO3

–. If an ion is amphiprotic, it acts like an acid and a base. We must determine which reaction is predominant. Compare the Ka of the acid to the Kb of the base. The larger equilibrium constant will be the predominant reaction. Ex 2: Are HCO3

–, HSO3– acidic or basic?

A solution can form with a cation that is acidic, and an anion that is basic. The predominant reaction is determined by comparing the Ka and the Kb. Ex 3: Predict the acidity/basicity of: NH4NO2 b) NH4CN c) KHSO4 d) Al(NO3)3 Follow Up Problems 18.11, 12 Problems 18.117, 119, 121, 123, 127


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IB Chemistry ABS — Buffers/Indicators

A buffer is an application of Le Châtelier's Principle to acid/base equilibrium. A buffer is a mixture of a weak acid and its conjugate base that minimize the change in pH upon the addition of small amounts of strong acid or base. Buffers are very important in many biological systems. Consider an acetic acid buffer: CH3COOH + H2O H3O+ + CH3COO–

What happens with the addition of a small amount of strong acid? Strong base? Describe three ways to make this buffer.


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A buffer is a weak acid equilibrium. For: HA + H2O H3O+ + A–

If [HA] = [A–] then Ka = [H3O+] and pH = pKa The appropriate acid for a specific pH buffer can be chosen by pKa values. For a buffer, the pH can be calculated from the concentrations of the acid and the conjugate base.

Taking the -log of both sides: Ex 1: What is the pH of a buffer with 0.20 moles of H2CO3 and 0.35 moles of HCO3

–? What would happen to the pH is 0.010 moles of OH– is added to the buffer? Ex2: How would you prepare a buffer at a pH of 11.0?

Ka=

[H3O

+][A

-]

[HA]

Ka=

[H3O

+][A

-]

[HA]

[H3O

+]=K

a

[HA]

[A-]

pH = pKa + log[base]

[acid]


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The optimum pH of blood is 7.35. Haemoglobin in blood binds with oxygen through the following equilibrium: HHb + O2 + H2O H3O+ + HbO2

–

What happens if blood pH gets too high or low? Main buffers in blood are H2CO3/HCO3

– and H2PO4–/HPO4

2– An acid/base indicator is a weak acid/ conjugate base mixture where the acid and base have different colours. Bromophenol blue has a yellow and blue colour. HBrph + H2O H3O+ + Brph–

(yellow) (blue) With the addition of acid, the equilibrium shifts left, becomes yellow. With the addition of base, the equilibrium shifts right, becomes blue. In the middle of the colour change (green) [HBrph] = [Brph–]. Ka = [H3O+]

Indicators are added to titrations to

determine the endpoint of the titration. They are added in such small amounts that they do not affect the the titration. There is a table in your IB data booklet that outlines the pH range and colour change for different indicators. Follow Up Problems 19.1, 2 Problems 19.2, 4, 7, 9, 11, 13, 19, 27, 33, 39, 40, 41, 48

Ka=

[H3O+ ][Brph-]

[HBrph]


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IB Chemistry ABS — Titration curves The pH scale is logarithmic. One drop of HCl may change the pH of a solution from 7.0 to 5.0 (1.0x10-7M to 1.0x10-5M) yet have no measurable affect on a solution at a pH of 1.0 (0.1 M) During a titration the pH will change around the neutral solution very quickly. A strong acid/strong base titration will reach the equivalence point at pH = 7.0. A titration of a strong acid with a strong base (base in the burette): Similarly, a strong base titrated with a strong acid: An indicator for this titration must change in the range of pH 3 - 11. What indicators would be suitable?


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There are three important differences with weak acid/base titrations. 1. The initial pH will be different because the weak acid/base

only dissociates partially. 2. A buffer will form as part of the weak acid or base has been

converted into its conjugate. 3. The pH at the equivalence point will not be 7.0. At the

equivalence point, there is a conjugate acid/base to alter the pH.

To determine the pH at the equivalence point, remember the species that are present. A weak acid will be reacted to its conjugate base: the solution will be basic. A weak base will be reacted to its conjugate acid: the solution will be acidic. The indicator must be chosen accordingly.


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A titration of a weak acid with a weak base, will have two buffers. There will be no clear equivalence point, and the titration will not be useful Draw a weak acid/weak base titration curve The buffer region can be used to calculate the K value of the weak acid or base. If a weak acid titration used 20.00 mL of base to reach the equivalence point, then at 10.00 mL, half of the acid will be converted to the conjugate base.

Follow Up Problems 19.3(e) Problems 19.42, 44, 46, 50, 58 Sketch all titration curves for these problems

Ka=

[H3O

+][A

-]

[HA]K

a = [H

3O

+]

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IB Chemistry ABS — Introductionsemmchem.cmswiki.wikispaces.net/file/view/ibchabs.pdf/482713820/... · IB Chemistry — Acids Bases and Salts 1 IB Chemistry ABS — Introduction