Question

A 200 mL solution of $HCl$ having concentration of $2.5\times {10}^{-2}M$ is mixed with an another 300 mL solution of $NaOH$ having a concentration of 0.01 M. The resultant pH of the mixture is :

• A. 12.5
• B. 9.6
• C. 4.7
• D. 2.4

Answer 1

The correct option is D 2.4

Solution 1 :

$HCl\left(aq\right)\to H^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
Since HCl is a strong acid, all the HCl dissociates into its ions .
At equilibrium,
$[HCl{]}_{initially}=[H^{+}{]}_{equilibrium}=2.5\times {10}^{-2}M\therefore C_1=2.5\times {10}^{-2}M$
$V_1=200mL$

Solution 2 :

$NaOH\left(aq\right)\to N{a}^{+}(aq.)+O{H}^{-}\left(aq\right)$
Since NaOH is a strong base, it dissociates completely into its ions.
At equilibrium,
$[NaOH{]}_{initially}=[O{H}^{-}{]}_{equilibrium}=0.01M\therefore C_2=0.01M{V}_2=300mL$

$C_{mix}=\frac{C_1{V}_1-C_2{V}_2}{V_1+V_2}{C}_{mix}=\frac{(2.5\times {10}^{-2}\times 200)-(0.01\times 300)}{(200+300)}{C}_{mix}=4\times {10}^{-3}MpH=-log\left[C_{mix}\right]pH=-log(4\times {10}^{-3})=(3-log4)=(3-2log2)pH=3-0.6=2.4pH=2.4$