Ksp(BaSO4) is 1.1×10−10. In which case is BaSO4 precipitated?

100 mL of 4 × 10 − 3 M of BaCl 2 + 300 mL of 6.0 × 10 − 4 M of Na 2 SO 4

Ksp(BaSO4) is 1.1×10−10. In which case is BaSO4 precipitated?

100 mL of 4 × 10 − 3 M of BaCl 2 + 300 mL of 6.0 × 10 − 4 M of Na 2 SO 4

100 mL of 4 × 10 − 4 M of BaCl 2 + 300 mL of 6 × 10 − 8 M of Na 2 SO 4

100 mL of 4 × 10 − 4 M of BaCl 2 + 100 mL of 6.0 × 10 − 8 M of Na 2 SO 4

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$K_{sp} ({BaSO}_{4})$ is $1.1 \times 10^{- 10}$ . In which case is ${BaSO}_{4}$ precipitated?

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100 mL of $4 \times 10^{- 3} M$ of

${BaCl}_{2} + 300 mL of 6.0 \times 10^{- 4} M {of Na}_{2} {SO}_{4}$

100 mL of $4 \times 10^{- 4}$ M of

${BaCl}_{2} + 300 mL of 6 \times 10^{- 8} M {of Na}_{2} {SO}_{4}$

100 mL of $4 \times 10^{- 4} M$ of

${BaCl}_{2} + 100 mL of 6.0 \times 10^{- 8} M {of Na}_{2} {SO}_{4}$

In all cases

answer is A.

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Detailed Solution

Before mixing, $[B a^{+ 2}] = 4 \times 10^{- 3} M \Rightarrow [S O_{4}^{- 2}] = 6 \times 10^{- 4} M$
after mixing,
$\begin{array}{l} 4 \times 10^{- 3} \times 100 = M_{2} \times 400 \\ [B r^{+ 2}] = 10^{- 3} M \\ 6 \times 10^{- 4} \times 300 = M_{2} \times 400 \\ [S O_{4}^{- 2}] = 4.5 \times 10^{- 4} M \end{array}$
$I P = [B a^{+ 2}] [S O_{4}^{- 2}] = 4.5 \times 10^{- 7}; K s p < I . P .$