Question

$50mL$ of $H_{2}O$ is added to $50mL$ of $1\times {10}^{-3}M$ barium hydroxide solution. What is the $pH$ of the resulting solution?

• A. $3.0$
• B. $3.3$
• C. $11.0$
• D. $11.7$

Answer 1

Answer: (C)

$11.0$

Barium hydroxide is $Ba(OH{)}_2$

Since, one mole of $Ba(OH{)}_2$ gives $2$ $moles$ of $O{H}^{-}$ ions,

$\therefore$ Moles of $O{H}^{-}=2\times$ Molarity of $Ba(OH{)}_2\times$Volume $=2\times 1\times {10}^{-3}\times 50\times {10}^{-3}={10}^{-4}$ $mol$

Final volume $=50+50=100$ $mL={10}^{-1}$ $L$

$\therefore$ Final concentration of $O{H}^{-}=\left[O{H}^{-}\right]=\frac{{10}^{-4}}{{10}^{-1}}={10}^{-3}$

$\therefore pOH=-\log {10}^{-3}=3$

$\therefore pH=14-3=11$