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

Feb 17, 2017

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

Explanation:

BaSO_4(s)⇌Ba^(2+)(aq)+SO_4^(2-)(aq)

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

$\left[B {a}^{2 +}\right] = \left[S {O}_{4}^{2 -}\right]$ (As both are equal according to moles)

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

${K}_{s p} = \left[B {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] = {\left(0.0000105\right)}^{2} = 1.1 \times {10}^{-} 10$