Question

Consider two strong acid solutions : Solution A has a pH value of 2 while the pH value of solution B is 3. Solution A has twice the volume of solution B. The resultant pH after mixing of these two solutions is :
Take $\log \left(7\right)=0.84$

• A. 2.16
• B. 2.84
• C. 2.5
• D. 3.84

Answer 1

The correct option is A 2.16

pH of solution A = 2
We know that $pH=-log\left[H^{+}\right]$
$2=-log\left[H^{+}\right]$
$[H^{+}{]}_{SolutionA}={10}^{-2}M$

pH of solution B = 3
Since, $pH=-log\left[H^{+}\right]$
$3=-log\left[H^{+}\right]$
$[H^{+}{]}_{SolutionB}={10}^{-3}M$

If volume of solution B = V
Then, volume of solution A = 2V
$V_1=2V{V}_2=V$

$\therefore$ Total volume $\left(V_f\right)=V_1+V_2=2V+V$

$V_f=3V$
$C_{mixture}=[H^{+}{]}_{mixture}=\frac{C_1{V}_1+C_2{V}_2}{V_1+V_2}[H^{+}{]}_{mixture}=\frac{({10}^{-2}\times 2V)+({10}^{-3}\times V)}{3V}[H^{+}{]}_{mixture}=\frac{(2\times {10}^{-2})+\left({10}^{-3}\right)}{3}[H^{+}{]}_{mixture}=7\times {10}^{-3}M$

$p{H}_{mix}=-log[H^{+}{]}_{mixture}p{H}_{mix}=-log(7\times {10}^{-3})p{H}_{mix}=(3-log\left(7\right))=(3-0.84)p{H}_{mix}=2.16$