Question

Only litre buffer solution was prepared by adding 0.10 mol each of $ NH_3 $ and $ NH_4Cl $ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is _____ $ \times 10^{-2} $, (Nearest integer) (Given : $ pK_b $ of $ NH_3 = 4.745 $ and $ \log_{10}3 = 0.477 $)

Hint:

When dealing with buffer solutions, the Henderson-Hasselbalch equation is a quick and effective tool. Remember to account for the moles of acid or base added reacting with the buffer components to find the new concentrations before calculating the final pH. The change in pH is the difference between the final and initial pH values.

Answer 1

Correct Answer: 48

Step 1: Determine the initial pH of the buffer.
For a basic buffer, the initial pOH is calculated using the Henderson-Hasselbalch equation: $$pOH_{initial} = pK_b + \log_{10} \frac{[salt]}{[base]} = 4.745 + \log_{10} \frac{0.10}{0.10} = 4.745$$ The initial pH is then found by: $$pH_{initial} = 14 - pOH_{initial} = 9.255$$
Step 2: Calculate the pH after adding HCl.
The addition of HCl results in the following reaction, altering the concentrations of the base and its conjugate salt: $$NH_3 + HCl \rightarrow NH_4^+ + Cl^-$$ The new molar concentrations are: \( [NH_3] = 0.05 \) M, \( [NH_4^+] = 0.15 \) M. The final pOH is calculated as: $$pOH_{final} = pK_b + \log_{10} \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log_{10} \frac{0.15}{0.05} = 4.745 + 0.477 = 5.222$$ The final pH is: $$pH_{final} = 14 - pOH_{final} = 8.778$$ 
Step 3: Calculate the pH change.
The change in pH is determined by: $$\Delta pH = pH_{final} - pH_{initial} = 8.778 - 9.255 = -0.477$$ 
The absolute value of the pH change is \( |\Delta pH| = 0.477 \). 
Expressed in the specified format: \( 0.477 = 47.7 \times 10^{-2} \). 
Rounding to the nearest integer yields 48.