A buffer solution is made by mixing a weak acid HA $(K_{a} = 10^{- 6})$ with its salt $NaA$ in equal amounts. What should be amount of acid or salt that should be added to make 90 mL of buffer solution in which if 0.1 mole of strong acid are added into 1 L of this buffer solution then change in pH is unity?

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9 m moles

22 m moles

11 m moles

10 m moles

answer is D.

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Detailed Solution

$\begin{array}{l} A^{-} + H^{+} ⇌ HA \\ \begin{array}{ccc} a & b & a \\ a - b & - & a + b \end{array} \\ b = \frac{0.1 \times 90}{1000} = 9 \times 10^{- 3} \\ a = 11 m mol \\ pH i = p K_{a} \\ {pH}_{f} = p K_{a} + 1 \end{array} \begin{aligned} \Rightarrow \frac{a + b}{a - b} = 10 \\ \Rightarrow a = \frac{11 b}{9} \end{aligned}$