Question

Question: Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

Short answer

Answer

pH of 0.1 M CH₃NH₃Cl = 5.8

pH of 0.05M NaCN = 10.95

Step-by-step solution

Ka Calculation of CH3NH3Cl 

CH₃NH₃Cl gives CH₃NH₃⁺ and Cl^- on dissociation. Cl^- does not hydrolyse as it is a weak conjugate base of strong acid.

CH₃NH₃⁺ hydrolyses as:

${\text{CH}}_{\text{3}}\text{N}{{\text{H}}_{\text{3}}}^{\text{+}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

K_a can be calculated as

${\text{K}}_a=\frac{{\text{[K}}_{\text{w}}\text{]}}{{\text{[K}}_{\text{b}}\text{]}}=\frac{1\times {10}^{-14}}{4.38\times {10}^{-4}}=2.3\times {10}^{-11}$

 Step 2: pH Calculation of CH3NH3Cl 

 Step 3: Ka Calculation of NaCN

NaCN gives Na⁺ and CN^- on dissociation. Na+ does not hydrolyse .

CN^- hydrolyses as:

${\text{CN}}^{-}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{HCN + OH}}^{-}$

K_b can be calculated as

${\text{K}}_{\text{b}}=\frac{{\text{[K}}_{\text{w}}\text{]}}{{\text{[K}}_a\text{]}}=\frac{1\times {10}^{-14}}{6.2\times {10}^{-10}}=1.6\times {10}^{-5}$

 Step 4: pH Calculation of NaCN