Question

The Ksp of barium sulfate (BaSO4) is 1.1 × 10–10. What is the solubility concentration of sulfate ions in a saturated solution at 25°C?
5.5 × 10–11M
1.0 × 10–5M
7.4 × 10–6M
1.5 × 10–5M

Answer 1

You can find this just by looking at the Ksp equation for BaSO4. First of all, note that each molecule of BaSO4 dissociates into one Ba(2+) ion and one SO4(2-). The Ksp equation is then:

\[K_{sp}=[Ba^{2+}][SO_4^{2-}]\]
We know that for however much BaSO4 dissolves, the same amount of each ion is produced, meaning each also will have the same concentration - let's call that concentration x. We're given the value for Ksp in the question so:

\[1.1 \times 10^{10}=x^2\]\[x=1.05 \times 10^{-5}\]
That means the concentration of each ion, including sulfate, is 1.05 x 10^(-5) M at equilibrium! Option B is your answer!