It is observed that 7.5 mmol of BaF₂ will dissolve in 1.0 L of water. Use these data to calculate the value of K_sp for barium fluoride. A) 1.7 × 10^-6 B) 5.6 × 10^-5 C) 2.1 × 10^-12 D) 7.5 × 10^-3 E) 4.2 × 10^-7

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Quizplus · Accepted Answer

The solubility product constant, $K_{sp}$, for barium fluoride (BaF2) can be calculated using the dissolution equation: $BaF2 \rightarrow Ba^{2+} + 2F^-$. From the given solubility of 7.5 mmol/L, we can deduce that the concentration of $Ba^{2+}$ ions is 7.5 mmol/L and the concentration of $F^-$ ions is $2 \times 7.5$ mmol/L = 15 mmol/L. Therefore, $K_{sp} = [Ba^{2+}][F^-]^2 = (7.5 \times 10^{-3})(15 \times 10^{-3})^2 = 1.7 \times 10^{-6}$.

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