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 $)
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Correct Answer: 48
Approach Solution - 1
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.
Approach Solution -2
Initial moles: NH3 = 0.10, NH4+ (from NH4Cl) = 0.10. Volume ≈ constant, so Henderson–Hasselbalch (base buffer) applies.
pOHinitial = pKb + log([salt]/[base]) = 4.745 + log(0.10/0.10) = 4.745
pHinitial = 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
pOHfinal = pKb + log(0.15/0.05) = 4.745 + log 3 = 4.745 + 0.477 = 5.222
pHfinal = 14 − 5.222 = 8.778
ΔpH = pHinitial − pHfinal = 9.255 − 8.778 = 0.477 = 47.7 × 10−2 ≈ 48 × 10−2
Answer: 48
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