Question

What $\left[{OH}^{-}\right]$ in the final solution prepared by mixing $20mL$ of $0.05M$ $HCl$ with $30mL$ of $0.10M$ $Ba{\left(OH\right)}_2$?

• A. $0.12M$
• B. $0.10M$
• C. $0.40M$
• D. $0.005M$

Answer 1

The correct option is B $0.10M$

Number of milli-equivalents of $Ba(OH{)}_2=30\times 0.1\times 2=6$

Number of milli-equivalents of $HCl=20\times 0.05=1$

Number of milli-equivalents of $Ba(OH{)}_2$ left after the addition of $HCl$$=(6-1)=5$

Total volume $=(20+30)mL=50mL=0.05litre$

$5$ milli-equivalents or $0.005$ equivalents of $Ba(OH{)}_2$ are present in 0.05 litre

$\frac{0.005}{0.05}=0.1$ equivalents of $Ba(OH{)}_2$ are present in one litre
So, $\left[{OH}^{-}\right]=0.1M$