Class XII students were asked to prepare one litre of buffer solution of pH \(8.26\) by their Chemistry teacher. The amount of ammonium chloride to be dissolved by the student in \(0.2\) M ammonia solution to make one litre of the buffer is (Given : \(pK_b\) \((NH_3)\) = \(4.74\), Molar mass of \(NH_3\) = \(17\) g \(mol^{–1}\), Molar mass of \(NH_4Cl\) = \(53.5\) \(g\) \(mol^{–1}\))
- 53.5 g
- 72.3 g
- 107.0 g
- 126.0 g
The Correct Option is C
Solution and Explanation
To calculate the amount of ammonium chloride (\(NH_4Cl\)) required to prepare a buffer solution with pH 8.26 using 0.2 M ammonia (\(NH_3\)), we use the Henderson-Hasselbalch equation for a basic buffer:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Since we are dealing with ammonia, a basic solution, we need to convert \(pK_b\) to \(pK_a\):
\[ \text{p}K_a = 14 - \text{p}K_b = 14 - 4.74 = 9.26 \]
Now, substitute the values into the Henderson-Hasselbalch equation:
\[ 8.26 = 9.26 + \log \left( \frac{0.2}{[\text{NH}_4^+]}\right) \]
Solving for \([\text{NH}_4^+]\):
\[ 8.26 - 9.26 = \log \left( \frac{0.2}{[\text{NH}_4^+]} \right) \]
\[ -1 = \log \left( \frac{0.2}{[\text{NH}_4^+]} \right) \]
Anti-logarithm of both sides:
\[ 10^{-1} = \frac{0.2}{[\text{NH}_4^+]} \]
\[ 0.1 = \frac{0.2}{[\text{NH}_4^+]} \]
\[ [\text{NH}_4^+] = \frac{0.2}{0.1} = 2 \text{ M} \]
Thus, the concentration of ammonium ion \([\text{NH}_4^+]\) in the solution should be 2 M. To find out how many grams of \(NH_4Cl\) are needed to make this 2 M solution, calculate using the molar mass:
\[ \text{Moles of } NH_4Cl = 2 \text{ moles/litre} \times 1 \text{ litre} = 2 \text{ moles} \]
Molar mass of \(NH_4Cl\) is 53.5 g/mol, so the mass required:
\[ \text{Mass of } NH_4Cl = 2 \text{ moles} \times 53.5 \text{ g/mol} = 107.0 \text{ g} \]
Thus, the student needs to dissolve 107.0 g of ammonium chloride in the solution. Therefore, the correct answer is 107.0 g.
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