Question

What approximate volume of 0.40 M $Ba(OH{)}_2$ must be added to 50.0 mL of 0.30 M NaOH to get a solution in which the molarity of the OH ions is 0.50 M?

• A. 33 mL
• B. 66 mL
• C. 133 mL
• D. 100 mL

Answer 1

Answer: (A)

33 mL

Let us suppose that x mL of 0.40 M $Ba{\left(OH\right)}_2$ soon is required to be added.

Assuming that $Ba{\left(OH\right)}_2$ dissociates completely, number of millimoles of $O{H}^{-}$ in solution $=x\times 0.80$ mmol

(Since each formula unit of $Ba{\left(OH\right)}_2$ gives 2 $O{H}^{-}$ ions.)

Number of millimoles of $O{H}^{-}$ due to dissociation of NaOH $=50\times 0.30$ =15 mmol

Total number of millimoles of $O{H}^{-}$ in solution$=15+x\times 0.80$ mmol

Total number of millimoles of $O{H}^{-}$ required for 0.50 M solution $=(50+x)mL\times 0.50$ mmol/mL

$=25+x\times 0.50$ mmol

Hence, $15+0.80\times x=25+x\times 0.50$

$\Rightarrow 0.30\Rightarrow x=10$

$x=\frac{10}{0.30}=33.33$

Approximately 33 mL of the $Ba{\left(OH\right)}_2$ solution needs to be added.