Question

100 mL solution of $HCl$ having concentration of $0.06M$ is mixed with an another 400 mL solution of $NaOH$ having a concentration of $0.01M$. The resultant pH of the mixture is :

• A. 1.5
• B. 9.6
• C. 2.4
• D. 11.5

Answer 1

The correct option is C 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,
$[HCl{]}_{initially}=[H^{+}{]}_{equilibrium}=0.06M\therefore C_1=0.06M$
$V_1=100mL$

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,
$[NaOH{]}_{initially}=[O{H}^{-}{]}_{equilibrium}=0.01M\therefore C_2=0.01M{V}_2=400mL$

$C_{mix}=\frac{C_1{V}_1-C_2{V}_2}{V_1+V_2}{C}_{mix}=\frac{(0.06\times 100)-(0.01\times 400)}{(100+400)}{C}_{mix}=4\times {10}^{-3}MpH=-log\left[C_{mix}\right]pH=-log(4\times {10}^{-3})=(3-log4)=(3-2log2)pH=3-0.6=2.4pH=2.4$