PS04 Acid Base - Problem set PDF

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UBC CHEM 123

PS04: Acid-Base Chemistry

Page 1 of 8

For all questions, assume that the given values are good to 4 significant figures. 1. For each of the following reactions, identify the role each chemical is playing: Bronsted-Lowry Acid, Bronsted-Lowry Base, conjugate acid, or conjugate base. Identify the conjugate acid-base pairs.

UBC CHEM 123

PS04: Acid-Base Chemistry

Page 2 of 8

2. (a) Acetic acid, CH3COOH, is a weak acid with a Ka of 1.8 x 10-5 at T = 25°C. Write the products of the acid-base reaction between acetic acid and water.

(b) All species in an equilibrium reaction are present (in varying amounts) in the same flask at equilibrium. The equilibrium distribution of products and reactants is quantified by the equilibrium constant, K, which depends on the reaction and reaction conditions. Write out the expression (in terms of concentrations) for the equilibrium constant, K, of the reaction from part (a).

K=

The Ka is an equilibrium constant for a particular type reaction. It applies to acid dissociation reactions where the equilibrium reaction is written to produce H3O+ in solution. (c) It is a bit more convenient to communicate (and compare) Ka values as pKa values. Calculate the pKa value of acetic acid.

pKa = –log(Ka) =

(d) The pH of a solution depends on the concentration of the hydronium ion, [H3O+], and is calculated as pH = –log[H3O+]. Aqueous solutions with pH < 7 are acidic and those with pH > 7 are basic. The useful range of the pH scale for aqueous solutions is from approximately 0 to 14. A 0.10 M aqueous solution of acetic acid has a pH of 2.87. Calculate the equilibrium concentrations of H3O+ and CH3COO‾. Hint: assume that [acetic acid]eq = [acetic acid]initial – x  [acetic acid]initial.

UBC CHEM 123

PS04: Acid-Base Chemistry

Page 3 of 8

3. (a) Ammonia, NH3, is a weak base with Kb = 1.8 x 10-5 at T = 25°C. (Note: that the Kb of ammonia and the Ka of acetic acid have the same numerical value. This is purely coincidental.) Label the acid and base on the reactants side, and the conjugate acid (CA) and conjugate base (CB) on the products side.

(b) Calculate the value of pKb of ammonia. pKb = (c) Calculate the [OH⁻] at equilibrium in a 1.0 M aqueous solution of ammonia.

(d) The pOH is analogous to the pH, and is calculated as pOH = –log[OH⁻]. Calculate the pOH of a 1.0 M aqueous solution of ammonia at equilibrium.

(e) In aqueous solution, the pH and pOH values are related as pH + pOH = 14. This is due to the autoionization of water with the equilibrium constant, Kw = 1 x 10-14. (More on this in a minute.) Calculate the pH of a 1.0 M aqueous solution of ammonia at equilibrium.

(f) Which of the following is true about a 1.0 M solution of NH3(aq) at equilibrium? Circle your answer. A. B. C. D. E.

[H3O+] [NH4+] [OH ] [NH4+] [NH3]

> = > > >

[OH ] 1.0 M 1.0 M [OH ] [NH4+]

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Page 4 of 8

PS04: Acid-Base Chemistry

4. (a) Hydrocyanic acid, HCN, is a weak acid with pKa = 9.2. Write out the equilibrium reaction between HCN and water. Label the acid, base, conjugate acid, and conjugate base.

(b) Potassium cyanide, KCN, is an ionic compound that dissociates completely to produce K+ ions and CN⁻ ions in aqueous solution. In aqueous solution, the potassium cation does not react with water (it is a spectator ion), but the cyanide anion does. Write the equilibrium reaction between the cyanide anion and water. Label the acid, base, conjugate acid and conjugate base.

(c) Is a 0.10 M aqueous solution of KCN acidic, neutral, or basic? Circle your answer. Acidic

Neutral

Basic

(d) For a conjugate acid-base pair, the Ka and Kb are related by KaKb = Kw, where Kw = 1.0 x 10-14. Calculate the Kb of CN⁻. This Kb value applies to the reaction from part (b).

(e) Calculate the pH of a 1.36 M aqueous solution of KCN.

5. Strong acids and strong bases dissociate 100% in water. There are six strong acids and 12 strong bases that you are expected to be able to recognize on sight as “strong” in Chem 123. The six strong acids are: HCl, HBr, HI, HClO4, HNO3, and H2SO4. The 12 strong bases are: NaOH, KOH, Ca(OH)2, etc (the hydroxides of the first two columns of the periodic table; excludes hydrogen). (a) Calculate the pOH and the pH of a 0.34 M solution of Mg(OH)2.

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Page 5 of 8

PS04: Acid-Base Chemistry

(b) Consider the reaction between 0.54 moles of HCl and 0.31 moles of KOH in a total volume of 1.00 L of aqueous solution. Since both HCl and KOH are strong acids, we know they will dissociate 100% in solution to give ions: HCl + H2O(l) → H3O+(aq) + Cl⁻(aq) KOH(s) + H2O(l) → K+(aq) + OH⁻(aq) Acids and bases present in the same solution will react to neutralize each other (usually producing water) until the supply of one of them runs out. (Recall limiting-reagent type questions from previous chemistry courses where the amounts given and reaction stoichiometry was used to determine the maximum amount of product that could be produced. Acid-base questions have similarities, but are usually interested in how much of the other reactant is leftover.) In this example, K+ and Cl⁻ are spectator ions. We can determine which species are present at equilibrium by setting up an “ICE” table (“ICE” stands for “initial”, “change”, and “equilibrium”, three distinct parts of a reaction. Sometimes they are called “RICE” tables, where “reaction” is included in the name.) ICE tables can be set up in moles or concentrations (molarities, M). It is important to be consistent with units within one ICE table. Complete the ICE table by filling in the two boxes with the equilibrium amounts. R

H 3O +

I

0.54 mol

0.31 mol

-

C

-0.31 mol

-0.31 mol

-

E

+

OH⁻

→

2 H 2O

-

(c) Calculate the pH of the resulting solution from part (b) (after the reaction has reached equilibrium).

(d) Consider the reaction between 1.28 g of hydroiodic acid, HI, and 0.400 g of sodium hydroxide, NaOH, in a total volume of 2.50 L of aqueous solution. Which species are present at equilibrium? What is the pH of the resulting solution?

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PS04: Acid-Base Chemistry

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6. Consider the following salts. Assume the ionic bond will break so that the salt dissociates completely into a solution of ions. Then, for each ion, consider whether it will undergo an acid-base reaction with water, or if it will be a spectator ion. The conjugate acids of strong bases, or the conjugate bases of strong acids do not react with water; they are spectator ions. Predict whether the resulting salt solution will be acidic, neutral, or basic (consider both ions).

Salt

Aqueous ions

Either write out the acid-base equilibrium (react the ion with water), or write “spectator ion”

pH of resulting solution (Circle one) Acidic

NaCN

Neutral Basic Acidic

KOOCCH3

Neutral Basic Acidic

NH4Cl

Neutral Basic Acidic

Na2SO4

Neutral Basic

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Page 7 of 8

PS04: Acid-Base Chemistry

7. (a) What is the pH of the solution formed by dissolving 16.05 g NH4Cl into 1.0 L water? The Ka of NH4+ is 5.56 x 10-10.

(b) A small amount (0.08 mol) of KOH is added to the solution from part (a). Which species react? Write out the acid-base reaction.

+

(c) What are the equilibrium concentrations of NH4 and NH3 after the KOH was added in part (b)? (Hint: complete this ICE table. Watch units.) R

+

⇌

I C E (d) What is the pH of the resulting solution?

(e) A buffer is a solution of a weak acid and its conjugate base mixed in appreciable amounts. A buffer is effective at resisting changes in pH when a small amount of strong acid or base is added. The effective range of a buffer solution is pH = pKa +/- 1 (i.e. the pH of the buffer solution has to be within 1 pH unit of the pKa). When strong acid (base) is added, it reacts with the weak base (acid) in solution so that there is not a large increase of [H3O+] ([OH⁻]) or decrease (increase) of pH. Is the solution formed in part (b) a buffer? Circle your answer:

Yes

No

UBC CHEM 123

PS04: Acid-Base Chemistry

Page 8 of 8

8. (a) A buffer solution is prepared by mixing together 2.17 mol CH3COOH and 1.83 mol CH3COONa in 0.500 L of water. The Ka of acetic acid is 1.8 x 10-5. What is the pH of the solution? Hint: consider using the Henderson-Hasselbalch equation here: pH = pKa + log([base]/[acid])

(b) A small amount (0.045 mol) of NaOH is added to the solution described in part (a). Neglect any changes in volume. Calculate the pH. Show your work.

(c) A small amount (0.09 mol) of HCl is added to the solution described in part (a). Neglect any changes in volume. Calculate the pH. Show your work.

(d) How many moles of HCl would have to be added to the solution described in part (a) for the resulting solution to have pH = 3.0? (Neglect any change in volume.) Show your work....