Question

What is the molarity of $K ^{+}$in aqueous solution that contains $17.4\,ppm$ of $K _{2} SO _{4}$ (molar mass $=174\, g\, mol ^{-1}$ )?

• A. $2 \times 10^{-2} M$
• B. $2 \times 10^{-3} M$
• C. $4 \times 10^{-4} M$
• D. $2 \times 10^{-4} M$

Answer 1

Answer: (D)

$K _{2} SO _{4}$ is $17.4\,ppm$, i.e. $10^{6} g$ has $17.4\, g\, K _{2} SO _{4}$
$1\, L \left(10^{3} mL \right)$ has $K _{2} SO _{4}$
$=\frac{17.4 \times 10^{3}}{10^{6}}=0.0174\, g / L$
$=\frac{0.0174}{174} mol / L$
$\therefore \left[ K _{2} SO _{4}\right]=1 \times 10^{-4} M$
$K _{2} SO _{4} \rightleftharpoons 2 K ^{+}+ SO _{4}^{2-}$
$\therefore \left[ K ^{+}\right]=2 \times 10^{-4} M$