1
Question
The solubility of barium sulphate at 298 K is 1.05×10mol dm. Calculate the solubility product.
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Solution
\(BaSO_{4(s)}\rightleftharpoons Ba^{2+}_{(aq)}+SO_{4(aq)}^{2-}\) Notice that each mole of barium sulphate dissolves to give 1 mole of barium ions and 1 mole of sulphate ions in solution. This means:
[Ba]=1.05×10 mol dm2+
[SO]=1.05×10 mol dm2−
All we need to do is to put these values into the solubility product expression, and do the simple sum.
Ksp=[Ba2+][SO2−4]
=(1.05×10−5)×(1.05×10−5)
=1.10×10−10 mol2 dm−6
Notice that each mole of barium sulphate dissolves to give 1 mole of barium ions and 1 mole of sulphate ions in solution.
This means:
[Ba]=1.05×10 mol dm2+
[SO]=1.05×10 mol dm2−
All we need to do is to put these values into the solubility product expression, and do the simple sum.
Ksp=[Ba2+][SO2−4]
=(1.05×10−5)×(1.05×10−5)
=1.10×10−10 mol2 dm−6
[Ba]=1.05×10 mol dm2+
[SO]=1.05×10 mol dm2−
All we need to do is to put these values into the solubility product expression, and do the simple sum.
Ksp=[Ba2+][SO2−4]
=(1.05×10−5)×(1.05×10−5)
=1.10×10−10 mol2 dm−6
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