What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2.0) is mixed with 300

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)?

Accepted Answer

$\left\lbrack H^{+} ightbrack = 10^{- 2}	ext{ M} \Rightarrow n_{H^{+}} = \frac{200 	imes 10^{- 2}}{1000} = 2 	imes 10^{- 3}$$\left\lbrack OH^{-} ightbrack = 10^{- 2}	ext{ M} \Rightarrow n_{OH^{-}} = \frac{300 	imes 10^{- 2}}{1000} = 3 	imes 10^{- 3}$∴ Moles of excess OH– remained = 1 × 10–3$\left\lbrack OH^{-} ightbrack = \frac{1 	imes 10^{- 3}}{500} 	imes 1000 = 2 	imes 10^{- 3}	ext{ M} $$$	herefore P^{OH} = - \log\left( 2 	imes 10^{- 3} ight) = 2.7 \Rightarrow P^{H} = 11.3$$

Question

Options

Solution

More Ionic Equilibrium Questions