Question

Solutions A and B both containing strong bases have pH= 12 and pH=13 respectively. Both the solutions have an equal volume. The pH of the resultant solution obtained after mixing the two solutions is:
Take $log\left(5.5\right)=0.74$

• A. 13.74
• B. 12.74
• C. 11.26
• D. 10.26

Answer 1

The correct option is B 12.74

pH of solution A = 12
$pOH=14-pHpOH=2$
We know that $pOH=-log\left[O{H}^{-}\right]$
$2=-log\left[O{H}^{-}\right]$
$[O{H}^{-}{]}_{1^{st}Solution}={10}^{-2}M$

pH of solution B = 13
$pOH=14-pHpOH=14-13=1pOH=1$
Since, $pOH=-log\left[O{H}^{-}\right]$
$1=-log\left[O{H}^{-}\right]$
$[O{H}^{-}{]}_{2^{nd}Solution}={10}^{-1}M$

Assuming both the solutions to be of $1L$
$V_1=V_2=1L$

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

$V_f=1+1=2L$

$C_{final}=[O{H}^{-}{]}_{final}=\frac{C_1{V}_1+C_2{V}_2}{V_1+V_2}[O{H}^{-}{]}_{final}=\frac{({10}^{-2}\times 1)+({10}^{-1}\times 1)}{2}[O{H}^{-}{]}_{final}=5.5\times {10}^{-2}M$

$pO{H}_{mix}=-log[O{H}^{-}{]}_{final}pO{H}_{mix}=-log(5.5\times {10}^{-2})pO{H}_{mix}=(2-log\left(5.5\right))=(2-0.74)pO{H}_{mix}=1.26p{H}_{mix}=14-1.26=12.74$