Question

What is the $\left[O{H}^{-}\right]$ in the final solution prepared by mixing $20.0$ mL of $0.050$ M HCl with $30.0$ mL of $0.10$ M $Ba(OH{)}_2$?

• A. $0.12$M
• B. $0.10$M
• C. $0.40$M
• D. $0.0050$M

Answer 1

Answer: (B)

$0.10$M

$Ba{\left(OH\right)}_2+2HCl⟶Ba{Cl}_2+2H_{2}O$

No. of moles of $H^{+}$ ions present in solution = concentration of $HCl\times$ Volume of $HCl$

$\Rightarrow 0.05\times \frac{20}{1000}=0.001\text{moles}$

There are two ${OH}^{-}$ ions in one molecule of $Ba{\left(OH\right)}_2$.

${OH}^{-}$ ions released in the Barium Hydroxide solution = concentration of $Ba{\left(OH\right)}_2\times$ Volume of $Ba{\left(OH\right)}_3\times 2$

$\Rightarrow \left[{OH}^{-}\right]=0.1\times \frac{30}{1000}\times 2=0.006\text{moles}$

$\because H^{+}$ ion is limiting reagent.

$\therefore$ 0.001 moles of $H^{+}$ and ${OH}^{-}$ combine to form water.

$\therefore$ 0.005 moles of ${OH}^{-}$ ions are remaining.

Total volume of solution $=(20+30)mL=50mL=0.05L$

$\therefore$ The molarity of ${OH}^{-}ion=\frac{0.005\text{moles}}{0.05L}=0.100M$