Question

Calculate the pH at 25^oC of the following aqueous salt solutions and indicate if they are basic, acidic, or neutral.

a) 0.020 M KClO₄(aq)

b)  0.50 M NaCN(aq)

c) 0.80 M NH₄ClO₄(aq)

Show all work please!

Answer 1

In solution, KClO₄dissociates completely to K⁺and ClO₄^- ,the ClO₄^-causes hydrolysis forming HClO4 and OH^-

ClO₄^- + H₂O $\rightleftharpoons$ HClO₄ + OH^-

therefore, we need the Kb of ClO4-

the Kb of HClO4 = 1 x 10^-21

Kb = [HClO₄] [OH^-] / [ClO₄^-]

$\Rightarrow$ 1 x 10^-21  = x² / (0.02 - x)

$\Rightarrow$ 1 x 10^-21  = x² (0.02

$\Rightarrow$ x² = 2 x 10^-23

$\Rightarrow$ x = 4.4 x 10^-12

[OH-] = x = 4.4 x 10^-12 M

pOH = -log (4.4 x 10^-12) = 11.36

pH = 14 - pOH = 14 - 11.36 = 2.64 (Acidic)

(b)

CN^- + H₂O $\rightleftharpoons$ HCN + OH^-

Kb = [HCN] [OH^-] / [CN^-] = 2.1 x 10^-5

Let
[HCH] = s
[OH-] = s
so, [CN-] = 0.5 - s

2.1 x 10^-5 = (s)(s) / (0.5 - s)

Assume s << 0.011

2.1 x 10^-5 = (s)(s) / (0.5)

s² = 1.05 x 10^-5

s = 3.24 x 10^-3 M

[OH-] =s = 3.24 x 10^-3 M

pOH = - log (3.24 x 10^-3) = 2.49

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 2.49 = 11.51 (Basic)

(c)

NH₄ClO₄ $\rightleftharpoons$ NH₄⁺ + ClO₄^-

Now in a water solution,
ClO₄^- + H₂O $\rightleftharpoons$ HClO₄ + OH^- (This doesn't happen because HClO₄ is a strong acid. Therefore the only contribution to the pH is from the NH₄⁺.

NH₄⁺    $\rightleftharpoons$    NH₃ + H⁺

IC 0.80 0 0

EC 0.80-x x x

Ka =  K_w/K_b = (1 x 10^-14) / (1.8 x 10^-5) = 5.5 x 10^-10

Ka = [NH₃] [H⁺] / [NH₄⁺]

$\Rightarrow$ 5.5 x 10^-10 = (x)(x)/(0.80-x)

$\Rightarrow$ ​5.5 x 10^-10 = x² /0.80 (Assume x << 0.80)

$\Rightarrow$ ​ x² = 4.4 x 10^-10

$\Rightarrow$ ​ x = 2.1 x 10^-5

[H⁺] = x = 2.1 x 10^-5

pH = -log(2.1 x 10^-5) = 4.68 (Acidic)