Question

Consider a strong base, $AOH$ solution having a $pH=11$ mixed in an equal volume with another solution of strong base, $BOH$ to give a resultant solution of $pH=12$. Find the $pH$ of solution containing strong base $BOH$.

• A. 13.9
• B. 12.8
• C. 12.3
• D. 13.3

Answer 1

The correct option is C 12.3

Solution 1 having strong base AOH:
$pH=11pOH=14-pHpOH=14-11=3pOH=3-log\left[O{H}^{-}\right]=3\left[O{H}^{-}\right]={10}^{-3}{C}_1={10}^{-3}M$
Solution 2 containing strong base BOH:
Let concentration be $C_2=\left[O{H}^{-}\right]$
Final solution obtained after mixing:
$pH=12pOH=14-pHpOH=14-12=2pOH=2-log\left[O{H}^{-}\right]=2\left[O{H}^{-}\right]={10}^{-2}M$

Concentration $\left(C_f\right)={10}^{-2}M$
The volume of two solutions are equal.
So,
$V_1=V_2=V{V}_f=V_1+V_2=2V{C}_f{V}_f=C_1{V}_1+C_2{V}_2{10}^{-2}\times 2V=({10}^{-3}\times V)+(C_2\times V)C_2=(2\times {10}^{-2})-\left({10}^{-3}\right)C_2=\left[O{H}^{-}\right]=0.019MpOH=-log\left[O{H}^{-}\right]pOH=-log\left(0.019\right)pOH\approx -log\left(0.02\right)pOH=2-log2=(2-0.3)=1.7pH=14-pOH=14-1.7=12.3pH=12.3$