Q. What is molarity of $K^{+}$ in aqueous solution that contains 17.4 ppm of $K_{2}{SO}_{4}$ ($174\ g\ {mol}^{- 1}$)?
A) $2 \times 10^{- 2}\ M$
B) $2 \times 10^{- 3}\ M$
C) $4 \times 10^{- 4}\ M$
D) $2 \times 10^{- 4}\ M$
Explanation
(d)
$K_{2}{SO}_{4}$ is 17.4 ppm i.e.
$10^{6}$ g ($\cong$mL)has $K_{2}{SO}_{4}$ = 17.4 g $K_{2}{SO}_{4}$
${\ \ 10}^{3}$ mL has $K_{2}{SO}_{4} = \frac{17.4\ \times 10^{3}}{10^{6}}$ = 0.0174 g / L
=$\frac{0.0174}{174}$ mol/L
$\therefore\ \lbrack K_{2}\ SO_{4}\rbrack\ \ = \ 1 \times \ 10^{- 4}\ \ M$
$K_{2}\ SO_{4}\ \ \rightleftharpoons 2K^{+}\ + \ SO_{4}^{2 -}$$\therefore\ \ \lbrack K^{+}\rbrack\ = \ 2\ \times 10^{- 4}\ M$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.