75. Calculate the pH of each of the following solutions: Use Ka for HCN = 4 x 10 -10 and Kb for (CH3)2NH = 7.4 x 10 -4 (a) 0.4735 M NaCN (b) 0.333 M [(C

Question

75. Calculate the pH of each of the following solutions: Use Ka for HCN = 4 x 10 -10 and Kb for (CH3)2NH = 7.4 x 10 -4
(a) 0.4735 M NaCN (b) 0.333 M [(CH3 )2 NH2 ]2 SO4

Answer 1

Hi Navarro

 

Before you answer this question you first need to determine whether NaCN and [(CH3)2NH2]2SO4 produce acidic or basic solutions. The salt in part (a) should produce a basic solution since CN-, the conjugate base of HCN is a strong base. On the other hand the salt in part (b) should produce an acidic solution since (CH₃)₂NH₂⁺, the conjugate acid of (CH₃)₂NH is a strong acid.

 

part (a)

 

First write down the equilibrium reaction when NaCN dissociates in water. This is

 

CN^- <=> HCN + OH^-

From the equilibrium reaction you should notice that this gives you an equilibrium expression for K_b. To find K_b use the following formula K_a*K_b = K_w. Solving for K_b gives

K_b = K_w/K_a = 1.0*10^-14/4*10^-10 = 2.5*10^-5.

Now you need to find the concentration of hydroxide ions in the solution. I don't know how your textbook shows it but I recommend to set up an ICE table.

 

    CN^-      <=> HCN + OH^-

I:  0.4735        0         0

C: -x               +x      +x

E: 0.4735-x     x          x

 

K_b = 2.5*10^-5= [HCN][OH^-]/[CN^-] = (x)(x)/0.4735-x = x²/0.4735

Solving for x gives you

x = [OH^-] = 3.4*10^-3 M  (check that this value is valid by the 5% rule)

pOH = -log[OH^-] = 2.47

pH = 14 - 2.47 = 11.5  As you can see NaCN forms a basic solution

 

part (b)

The salt forms an acidic solution so you need to find K_a.

K_a*K_b = K_w

Solving for K_a gives K_a = 1.0*10^-14/7.4*10^-4 = 1.4*10^-11

As in part a set up an ICE table to find the equilibrium concentrations.

 

    (CH₃)₂NH₂⁺ <=> (CH₃)₂NH + H⁺

I:  0.666                 0                0

C: -x                      +x              +x

E: 0.666-x               x                x

 

K_a = 1.4*10^-11 = [(CH₃)₂NH][H⁺]/[(CH₃)₂NH₂⁺] = (x)(x)/0.666-x = x²/0.666

Solving for x gives you

x = [H⁺] = 3.1*10^-6 M (check that this value is valid by the 5% rule)

pH = -log[H⁺] = 5.51  As you can see the salt forms an acidic solution

 

If you're having trouble identifying whether salts make neutral, acidic or basic solutions I suggest you consult your textbook.

 

Hope this answers your question.