Ionic EquilibriumHard
Question
What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2.0) is mixed with 300 ml of an aqueous solution of NaOH (pH = 12.0)?
Options
A.2.7
B.11.3
C.3.7
D.10.3
Solution
$\left\lbrack H^{+} \right\rbrack = 10^{- 2}\text{ M} \Rightarrow n_{H^{+}} = \frac{200 \times 10^{- 2}}{1000} = 2 \times 10^{- 3}$
$\left\lbrack OH^{-} \right\rbrack = 10^{- 2}\text{ M} \Rightarrow n_{OH^{-}} = \frac{300 \times 10^{- 2}}{1000} = 3 \times 10^{- 3}$
∴ Moles of excess OH– remained = 1 × 10–3
$\left\lbrack OH^{-} \right\rbrack = \frac{1 \times 10^{- 3}}{500} \times 1000 = 2 \times 10^{- 3}\text{ M} $$$\therefore P^{OH} = - \log\left( 2 \times 10^{- 3} \right) = 2.7 \Rightarrow P^{H} = 11.3$$
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