What is the molarity of K⁺ in aqueous solution that contains 17.4 ppm of K₂SO₄ (molar mass = 174 g mo^-1) ?

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$2 \times 10^{- 2} M$

$2 \times 10^{- 3} M$

$2 \times 10^{- 4} M$

$4 \times 10^{- 4} M$

answer is D.

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Detailed Solution

$K_{2} {SO}_{4} is 17 .4 ppm, i.e. 10^{6} g has 17 .4 {gK}_{2} {SO}_{4}$
$\begin{aligned} 1 L (10^{3} mL) has K_{2} {SO}_{4} = \frac{17 .4 \times 10^{3}}{10^{6}} = 0 .0174 g / L \\ = \frac{0 .0174}{174} mol / L \end{aligned}$
$\begin{array}{l} ∴ [K_{2} {SO}_{4}] = 1 \times 10^{- 4} M \\ K_{2} {SO}_{4} ⇌ 2 K^{+} + {SO}_{4}^{2} \\ ∴ [K^{+}] = 2 \times 10^{- 4} M \end{array}$