chemistry:solved problems 8

1.  Lactic acid (CH₃CHOHCOOH) has one acidic hydrogen. A 0.10 M aqueous solution of lactic acid has a pH of 2.43.

Calculate the value of K_a for lactic acid.

lactic acid

CH₃CHOHCOOH(aq)   +   H₂O(l) W H₃O⁺ (aq)   +   CH₃CHOHCOO ^– (aq)

K_a = ([H₃O⁺ ][CH₃CHOHCOO ^– ])/[CH₃CHOHCOOH]

[H₃O⁺ ]_eq = 10 ^{– pH} = 10 ^{– 2.43} = 0.0037 M

[CH₃CHOHCOO ^–]_eq = [H₃O⁺ ]_eq = 0.0037 M

[CH₃CHOHCOOH]_eq = 0.10 M – 0.0037 M = 0.10 M

K_a = [(0.0037)(0.0037)]/0.10 = 1.4 x 10 ^{– 4}

2.  A 0.020 M solution of niacin, one of the B vitamins, has a percent dissociation of 2.7%. Calculate the pH of  this solution

and the value of K_a for niacin.

niacin

C₅H₄NCOOH(aq)   +   H₂O(l) W H₃O⁺(aq)   +   C₅H₄NCOO ^– (aq)

K_a = ([H₃O⁺ ][C₅H₄NCOO ^– ])/[C₅H₄NCOOH]

We need to convert from a percent (0 – 100) scale to a fraction (0 – 1) scale.

2.7%/100% = 0.027

[H₃O⁺ ]_eq = 0.027(0.020 M) = 5.4 x 10 ^{– 4} M

pH = – log[H₃O⁺ ] = – log(5.4 x 10 ^{– 4})

pH = 3.27

[C₅H₄NCOO ^– ]_eq = [H₃O⁺ ]_eq = 5.4 x 10 ^{– 4} M

[C₅H₄NCOOH]_eq = 0.020 M – (5.4 x 10 ^{– 4} M) = 0.020 M

K_a = [(5.4 x 10 ^{– 4})(5.4 x 10 ^{– 4})]/0.020 = 1.5 x 10 ^{– 5}

3.  Calculate the pH and % dissociation of a benzoic acid (C₆H₅COOH) solution in which 1.53 g of benzoic acid is dissolved

in 250 mL of solution (K_a for benzoic acid is 6.3 x 10^-5).

benzoic acid

C₆H₅COOH(aq)   +   H₂O(l) W H₃O⁺(aq)   +   C₆H₅COO ^– (aq)

K_a = ([H₃O⁺ ][C₆H₅COO ^– ])/[C₆H₅COOH] = 6.3 x 10 ^{– 5}

We need to calculate the molarity of the benzoic acid using the following flow diagram.

mass ® moles ® molarity

The molar mass of benzoic acid is 122.12 g/mol.  Also, 250 mL equals 0.25 L.

1.53 g (1 mol/122.12 g/mol) = 0.0125 mol

0.0125 mol/0.25 L = 0.050 M

[C₆H₅COOH]                         [H₃O⁺]                   [C₆H₅COO ^– ]

Initial                             0.050                                     0                                 0

Change                            – x + x + x

Equilibrium         0.050 – x x x

Since 100(K_a) < 0.050 [0.0063 < 0.050] Þ 0.050 – x . 0.050

6.3 x 10 ^{– 5} = x²/0.050 Þ x² = 3.2 x 10 ^{– 6}

x = 0.0018

ˆ [H₃O⁺ ] = 0.0018 M

pH = – log[H₃O⁺ ] = – log(0.0018)

pH = 2.75

% dissociation = ([H₃O⁺ ]_eq/[C₆H₅COOH]_initial) x 100%

% dissociation = (0.0018M/0.050M) x 100%

% dissociation = 3.6%

4.  Write an equation and the corresponding equilibrium-constant expression to describe the proton transfer reaction that

occurs when each of these bases is added to water.

(a)  C₆H₅N

C₆H₅N(aq)   +   H₂O(l) W C₆H₅NH⁺(aq)   +   OH ^– (aq)

K_b = ([C₆H₅NH⁺ ][OH ^– ])/[C₆H₅N]

(b)  HCO₃ ^–

HCO₃ ^– (aq)   +   H₂O(l) W H₂CO₃(aq)   +   OH ^– (aq)

K_b = ([H₂CO₃][OH ^– ])/[HCO₃ ^– ]

5.  Identify the major and minor (if present) species present in the following aqueous solutions.

(a) NH₃(aq) Þ Weak base

major species Þ NH₃

minor species Þ NH₄ ⁺ and OH ^–

(b) LiOH(aq) Þ Strong base

major species Þ Li ⁺ and OH ^–

no minor species present

6.  Calculate the pH of a 0.200 M solution of triethylamine, (CH₃CH₂)₃N (K_b = 4.0 x 10^-4).

triethyl amine

(CH₃CH₂)₃N(aq)   +   H₂O(l) W (CH₃CH₂)₃NH⁺(aq)   +   OH ^– (aq)

K_b = ([(CH₃CH₂)₃NH⁺][OH ^– ])/[(CH₃CH₂)₃N] = 4.0 x 10 ^{– 4}

[(CH₃CH₂)₃N]                        [(CH₃CH₂)₃NH ⁺]  [OH ^– ]

Initial                             0.200                                           0                      0

Change                            – x + x + x

Equilibrium           0.200 – x x x

Since 100(K_b) < 0.200 [0.040 < 0.200] Þ 0.200 – x . 0.200

4.0 x 10 ^{– 4} = x²/0.200 Þ x² = 8.0 x 10 ^{– 5}

x = 0.0089

ˆ [OH ^– ] = 0.0089 M

pOH = – log[OH ^– ] = – log(0.0089)

pOH = 2.05

pH = 14.00 – pOH

pH = 14.00 – 2.05 = 11.95

7.   Predict whether the following salts produce acidic, basic, or neutral solutions when dissolved in water. Explain your

reasoning in each case. You may need to refer to data found in Tables C1, C2, and C3 in Appendix II in the textbook.

(a) KCl

K⁺ Þ +1 monatomic cation Þ neutral

Cl ^– Þ conjugate base of the strong acid HCl Þ neutral

KCl produces a neutral solution when dissolved in water.

(b) NaCN

Na⁺ Þ +1 monatomic cation Þ neutral

CN ^– Þ conjugate base of the weak acid HCN Þ basic

NaCN produces a basic solution when dissolved in water.

(c) NH₄NO₃

NH₄⁺ Þ conjugate acid of the weak base NH₃ Þ acidic

NO₃ ^– Þ conjugate base of the strong acid HNO₃ Þ neutral

NH₄NO₃ produces an acidic solution when dissolved in water.

(d) AlBr₃

Al³⁺ Þ +3 cation Þ acidic

Br ^– Þ conjugate base of the strong acid HBr Þ neutral

AlBr₃ produces an acidic solution when dissolved in water.

(e) Cu(C₂H₃O₂)₂

Cu²⁺ Þ +2 transition metal cation Þ acidic

C₂H₃O₂ ^– Þ conjugate base of the weak acid HC₂H₃O₂ Þ basic

In order to determine whether this solution is acidic or basic we need to compare K_a for the    hydrated Cu²⁺ ion and K_b for the acetate ion.

K_a = 3 x 10^-8 (from Table C2 in Appendix II)

We can determine K_b for the acetate ion from knowledge of K_a for acetic acid.

Remember that for a conjugate acid-base pair we have the following relationship.

K_a x K_b = K_w

K_b = K_w/ K_a = (1.0 x 10^-14)/(1.8 x 10^-5) = 5.6 x 10^-10

Since the value of K_a is greater than the value of K_b this aqueous solution is acidic.

8.  Place the species in each of the following groups in order of increasing acid strength. Explain your reasoning in each

case.

You need to identify whether the acids in each group are binary acids or oxyacids.

(a)  HBrO, HBrO₃, HBrO₂ Þ oxyacids

HBrO < HBrO₂ < HBrO₃

Acid strength increases with the number of oxygen atoms in the acid.

(b)  HClO₂, HIO₂, HBrO₂ Þ oxyacids

HIO₂ < HBrO₂ < HClO₂

Acid strength increases with the electronegativity of element Y.

(c)  HI, HF, HBr  Þ binary acids

HF < HBr < HI

Acid strength increases as H-X bond strength decreases.