When 200 mL of aqueous solution of $HCl$ ($pH = 2$) is mixed with 300 mL of an aqueous solution of $NaOH$ ($pH = 12$), the pH of the resulting mixture is

Question

Dr. Abubacker · Accepted Answer

(d)
Concentration of HCl solution$= 1 	imes 10^{- 2}$
$	herefore$ Millimoles of HCl solution$= 200 	imes 1 	imes 10^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 2$
Similarly, millimoles of NaOH solution
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 300 	imes 1 	imes 10^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 3$
Concentration of the resultant solution
$= \frac{3 - 2}{300 + 200}$
$= \frac{1}{500} = 0.2 	imes 10^{- 2}$
$\left\lbrack OH^{-} ightbrack = 0.2 	imes 10^{- 2}$
$\left\lbrack H^{+} ightbrack = \frac{10^{- 14}}{0.2 	imes 10^{- 2}} = 5 	imes 10^{- 12}$
$pH = - \log{\lbrack H^{+}brack}$
$pH\  = - \log{\lbrack 5 	imes 10^{- 12}brack}$
$pH = \ 11.3$

Q. When 200 mL of aqueous solution of $HCl$ ($pH = 2$) is mixed with 300 mL of an aqueous solution of $NaOH$ ($pH = 12$), the pH of the resulting mixture is

Explanation

Reviewed by Dr. Abubacker

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