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: Calculate the initial pH of the buffer solution.
Using the Henderson-Hasselbalch equation for a basic buffer: $$pOH_{initial} = pK_b + \log_{10} \frac{[salt]}{[base]} = 4.745 + \log_{10} \frac{0.10}{0.10} = 4.745$$ $$pH_{initial} = 14 - pOH_{initial} = 9.255$$ 
Step 2: Calculate the pH after the addition of HCl.
The reaction with HCl changes the concentrations of the base and its salt: $$NH_3 + HCl \rightarrow NH_4^+ + Cl^-$$ New moles: \( [NH_3] = 0.05 \) M, \( [NH_4^+] = 0.15 \) M $$pOH_{final} = pK_b + \log_{10} \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log_{10} \frac{0.15}{0.05} = 4.745 + 0.477 = 5.222$$ $$pH_{final} = 14 - pOH_{final} = 8.778$$ 
Step 3: Calculate the change in pH.
$$\Delta pH = pH_{final} - pH_{initial} = 8.778 - 9.255 = -0.477$$ 
The magnitude of the change is \( |\Delta pH| = 0.477 \). 
Expressing this in the required format: \( 0.477 = 47.7 \times 10^{-2} \). 
Rounding to the nearest integer gives 48.

Answer 2

Initial moles: NH3 = 0.10, NH4⁺ (from NH4Cl) = 0.10. Volume ≈ constant, so Henderson–Hasselbalch (base buffer) applies.
pOH_initial = pK_b + log([salt]/[base]) = 4.745 + log(0.10/0.10) = 4.745
pH_initial = 14 − 4.745 = 9.255 

Add 0.05 mol HCl: NH3 + HCl → NH4⁺
New moles: NH3 = 0.10 − 0.05 = 0.05; NH4⁺ = 0.10 + 0.05 = 0.15
pOH_final = pK_b + log(0.15/0.05) = 4.745 + log 3 = 4.745 + 0.477 = 5.222
pH_final = 14 − 5.222 = 8.778

ΔpH = pH_initial − pH_final = 9.255 − 8.778 = 0.477 = 47.7 × 10⁻² ≈ 48 × 10⁻²
Answer: 48