Question

Upon mixing equal volume of two solution of strong acids having pH values $2$ & $4$,resulting solution will have pH:

• A. equal to 3
• B. closer to 2 than 4
• C. closer to 4 than 2
• D. closer to 3 than 2

Answer 1

The correct option is B closer to 2 than 4

Solution:- (B) closer to $2$ than $4$

As we know that,

$pH=-log\left[H^{+}\right]$

For the solution of $pH2$-

$\left[H^{+}\right]={10}^{-2}$

For the solution of $pH4$-

$\left[H^{+}\right]={10}^{-4}$

As given that equal volume of solutions are mixed.

Let the volume of both the solutions mixed be $V$ for each.

$\therefore$ Final volume of solution after mixing $\left(V_f\right)=V+V=2V$

Using formula-

$M_1{V}_1+M_2{V}_2=M_f{V}_f$

${10}^{-2}V+{10}^{-4}V=M_f\times 2V$

$\Rightarrow M_f=\frac{{10}^{-2}+{10}^{-4}}{2}$

$\Rightarrow M_f=0.505\times {10}^{-2}M$

$\therefore pH$ of the mixing solution will be-

${pH}_f=-\log 0.505\times {10}^{-2}$

$\Rightarrow {pH}_f=-(\log 0.505+\log {10}^{-2})$

$\Rightarrow {pH}_f=-0.297+2=1.703$

Hence the $pH$ of the solution after mixing will be closer to $2$.