Question

What is the molarity of $O{H}^{–}$ in the final solution prepared by mixing 20.0 mL of $0.10\text{M}$ $H_{2}S{O}_4$ with 30.0 mL of $0.10\text{M}$ $Ba(OH{)}_2$?

• A. $0.02\text{M}$
• B. $0.03\text{M}$
• C. $0.04\text{M}$
• D. $0.05\text{M}$

Answer 1

The correct option is C

$0.04\text{M}$

Millimoles of $H^{+}$ from $H_{2}S{O}_4$ $=\text{Molarity (M)}\times \text{Volume (mL)}\times \text{n-factor}\left(n_f\right)$
$=0.10\times 20\times 2=4\text{mmol}$

Millimoles of $O{H}^{-}$ from $Ba(OH{)}_2$ $=\text{Molarity (M)}\times \text{Volume (mL)}\times \text{n-factor}\left(n_f\right)$
$=0.1\times 30\times 2=6$mmol

$H^{+}+O{H}^{-}\to H_{2}O$
Initial mmoles: 4 6 0
After neutralization,
Moles of $O{H}^{-}$ left $=6-4=2\text{mmol}=2\times {10}^{-3}\text{mol}$

Molarity of $O{H}^{-}$ $=\frac{\text{Mole of solute}}{\text{Volume of solution in L}}$

$=\frac{2\times {10}^{-3}}{50\times {10}^{-3}}=0.04\text{M}$