Question

a. The pH of an aqueous solution of 0.272 M potassium hypochlorite, KClO (aq), is: __________...

a. The pH of an aqueous solution of 0.272 M potassium hypochlorite, KClO (aq), is: __________

is this solution acidic/basic/neutral?

b. The pH of an aqueous solution of 0.274 M sodium cyanide, NaCN (aq), is: __________

is this solution acidic/basic/neutral?

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Answer #1

a. The pH of an aqueous solution of 0.272 M potassium hypochlorite, KClO (aq), is: 10.445

is this solution acidic/basic/neutral? : basic

b. The pH of an aqueous solution of 0.274 M sodium cyanide, NaCN (aq), is: 11.416

is this solution acidic/basic/neutral? : basic

Explanation

(a.) initial concentration of ClO- = initial concentration of KClO = 0.272 M

Ka HClO = 3.5 x 10-8

Kb ClO- = (Kw) / (Ka)

Kb ClO- = (1.0 x 10-14) / (3.5 x 10-8)

Kb ClO- = 2.9 x 10-7

ICE table ClO- (aq) H2O (l) \rightleftharpoons HClO (aq) OH- (aq)
Initial conc. 0.272 M - 0 0
Change -x - +x +x
Equilibrium conc. 0.272 M - x - +x +x

Kb = [HClO]eq[OH-]eq / [ClO-]eq

2.9 x 10-7 = [(x) * (x)] / (0.272 M - x)

Solving for x, x = 2.786 x 10-4 M

[OH-] = x = 2.786 x 10-4 M

pOH = -log[OH-]

pOH = -log(2.786 x 10-4 M)

pOH = 3.555

pH = 14 - pOH

pH = 14 - 3.555

pH = 10.445

(b.) initial concentration of CN- = initial concentration of NaCN = 0.274 M

Ka HCN = 4.0 x 10-10

Kb CN- = (Kw) / (Ka)

Kb CN- = (1.0 x 10-14) / (4.0 x 10-10)

Kb CN- = 2.5 x 10-5

ICE table CN- (aq) H2O (l) \rightleftharpoons HCN (aq) OH- (aq)
Initial conc. 0.274 M - 0 0
Change -x - +x +x
Equilibrium conc. 0.274 M - x - +x +x

Kb = [HCN]eq[OH-]eq / [CN-]eq

2.5 x 10-5 = [(x) * (x)] / (0.274 M - x)

Solving for x, x = 2.786 x 10-4 M

[OH-] = x = 2.6 x 10-3 M

pOH = -log[OH-]

pOH = -log(2.6 x 10-3 M)

pOH = 2.584

pH = 14 - pOH

pH = 14 - 2.584

pH = 11.416

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