Question

The pH of a 0.02 MNH₄Cl solution will be

[Given K_b(NH₄ OH) = 10^-5 and log 2 = 0.301]
1. 2.6
2. 4.35
3. 4.65
4. 5.35

Answer 1

Correct Answer - Option 4 : 5.35

Concept:

pH is a measure of the amount of activity of hydrogen ions (H⁺) in a solution and thus its alkalinity or acidity.

pH is described as “the negative of the logarithm of the molar hydronium-ion concentration.”

The pH Formula can also be noted as pH = – log [H⁺]

Calculation:

The salt here is made from weak base and strong acid. So, the pH formula for this salt is given below:

\(\text{pH}=7-\frac{1}{2}\text{p}{{\text{K}}_{\text{b}}}-\frac{1}{2}\log \text{C}\)

∵ pK_b = – log K_b

\(\text{pH}=7-\frac{1}{2}\left( -\text{log}{{\text{K}}_{\text{b}}} \right)-\frac{1}{2}\log \text{C}\)     ----(1)          

Given that,

K_b (NH₄OH) = 10^-5,

Concentration, C = 0.02 = 2 × 10^-2 M

Now,

⇒ -log K_b = – log 10^-5

⇒ -log K_b = 5 log 10

∴ -log K_b = 5

Now,

⇒ log C = log (2 × 10^-2)

⇒ log C = log 2 + log 10^-2

⇒ log C = 0.301 + (-2 log 10)

⇒ log C = 0.301 – 2

∴ log C = - 1.699

Now, substituting the obtained values in equation (1),

\(\Rightarrow \text{pH}=7-\frac{1}{2}\left( 5 \right)-\frac{1}{2}\left( -1.699 \right)\)

\(\Rightarrow \text{pH}=7-\frac{5}{2}+\frac{1.699}{2}=\frac{14-5+1.699}{2}=\frac{10.699}{2}\)

∴ pH = 5.349 = 5.35