Barium sulfate, $B a S O_{4}$ $BaSO_4$ , is so insoluble that it can be swallowed without significant danger, even though $B a_{2}^{+}$ $Ba_2^+$ is toxic. At 25 °C, 1.00 L of water dissolves only 0.00245 g of $B a S O_{4}$ $BaSO_4$ . How do you calculate Ksp for $B a S O_{4}$ $BaSO_4$ ?

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Answer 1 · 2017-02-17T12:44:12

$1.1 \times 10^{- 10}$ $1.1 xx 10^-10$

$B a S O_{4} (s) ⇌ B a^{2 +} (a q) + S O_{4}^{2 -} (a q)$ $BaSO_4(s)⇌Ba^(2+)(aq)+SO_4^(2-)(aq)$

Moles of dissolved ${BaSO}_{4} = \frac{0.00245 g}{233.43 g/mol} = 0.0000105 mol$ $"BaSO"_4 = "0.00245 g" / "233.43 g/mol" = "0.0000105 mol"$ ;

$[B a^{2 +}] = [S O_{4}^{2 -}]$ $[Ba^(2+)]= [SO_4 ^(2-)]$ (As both are equal according to moles)

= $\frac{0.0000105 mol}{1.00 L} = 0.0000105 M$ $"0.0000105 mol"/"1.00 L" = 0.0000105 M$ ;

$K_{s p} = [B a^{2 +}] [S O_{4}^{2 -}] = {(0.0000105)}^{2} = 1.1 \times 10^{- 10}$ $K_(sp) = [Ba^(2+)][SO_4^(2-)]=(0.0000105)^2 = 1.1 xx 10^-10$

Barium sulfate, BaSO_4BaSO4BaSO_4, is so insoluble that it can be swallowed without significant danger, even though Ba_2^+Ba+2Ba_2^+ is toxic. At 25 °C, 1.00 L of water dissolves only 0.00245 g of BaSO_4BaSO4BaSO_4. How do you calculate Ksp for BaSO_4BaSO4BaSO_4?