A precipitate of strontium hydroxide Sr(OH)₂ (MM = 121.63 g mol⁻¹) was produced when 80.0 mL of 1.50 mol L⁻¹ strontium nitrate solution was mixed with 80.0 mL of 0.855 mol L⁻¹ sodium hydroxide solution - HSC - SSCE Chemistry - Question 35 - 2022 - Paper 1

Next, we need to find the initial concentrations of Sr²⁺ and OH⁻ in the mixed solution. For the strontium nitrate solution:

$n(Sr^{2+})_{initial} = 1.50 \, mol \, L^{-1} \times 0.0800 \, L = 0.120 \, mol$

For the sodium hydroxide solution:

$n(OH^{-})_{initial} = 0.855 \, mol \, L^{-1} \times 0.0800 \, L = 0.06840 \, mol$

At equilibrium, the concentration of Sr²⁺ is reduced by the moles of precipitate formed:

$[Sr^{2+}]_{equilibrium} = 0.120 \, mol - 0.03231 \, mol = 0.08769 \, mol$

To find the total volume, we add the volumes of both solutions:

$V_{total} = 80.0 \, mL + 80.0 \, mL = 160.0 \, mL = 0.1600 \, L$

Thus, the concentration of Sr²⁺ at equilibrium is:

$[Sr^{2+}]_{equilibrium} = \frac{0.08769 \, mol}{0.1600 \, L} = 0.5481 \, mol \, L^{-1}$

The concentration of OH⁻ at equilibrium is calculated similarly. The initial amount was:

$n(OH^{-})_{initial} = 0.06840 \, mol$

At equilibrium, after subtracting the precipitate formed:

$n(OH^{-})_{equilibrium} = 0.06840 \, mol - (2 \times 0.03231 \, mol) = 0.00378 \, mol$

Thus,

$[OH^{-}]_{equilibrium} = \frac{0.00378 \, mol}{0.1600 \, L} = 0.02363 \, mol \, L^{-1}$

Finally, the solubility product constant Ksp is calculated using the formula:

$K_{sp} = [Sr^{2+}][OH^{-}]^2$

Substituting the equilibrium concentrations:

$K_{sp} = (0.5481)(0.02363)^2 = 3.06 \times 10^{-4}$

Therefore, the Ksp of strontium hydroxide is:

$K_{sp} = 3.06 \times 10^{-4}$