AP Chemistry Unit 7.11: Ksp and Solubility

Practice AP Chemistry Unit 7.11 questions on writing Ksp expressions and calculating molar solubility, ion concentrations, or dissolved mass.

[Maximum number: 2]

Strontium hydroxide dissolves in water according to the following equation. The $K_{s p}$ expression for strontium hydroxide is provided.

\mathrm{Sr}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Sr}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{s p}=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}

Figure for Question 7.11 Introduction to Solubility Equilibria question 1 — AP Chemistry

(a)

The student prepares a saturated solution by adding excess $\mathrm{Sr}(\mathrm{OH})_{2}(s)$ to distilled water and stirring until no more solid dissolves. The student then determines that $\left[\mathrm{Sr}^{2+}\right]=0.043 M$ in the solution.

[ 2 ]

(i) For the correct calculated value: 1 point

\frac{0.043 \mathrm{~mol} \mathrm{Sr}^{2+}}{1 \mathrm{~L}} \times \frac{2 \mathrm{~mol} \mathrm{OH}^{-}}{1 \mathrm{~mol} \mathrm{Sr}^{2+}}=0.086 M \mathrm{OH}^{-}

(i)

Calculate the value of $\left[\mathrm{OH}^{-}\right]$ in the solution.

[ 1 ]

(i) For the correct calculated value: 1 point

\frac{0.043 \mathrm{~mol} \mathrm{Sr}^{2+}}{1 \mathrm{~L}} \times \frac{2 \mathrm{~mol} \mathrm{OH}^{-}}{1 \mathrm{~mol} \mathrm{Sr}^{2+}}=0.086 M \mathrm{OH}^{-}

(ii)

Calculate the value of $K_{s p}$ for $\mathrm{Sr}(\mathrm{OH})_{2}$ .

[ 1 ]

For the correct calculated value, consistent with (b)(i): 1 point

K_{s p}=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=(0.043)(0.086)^{2}=3.2 \times 10^{-4}

Total for part (b) 2 points

All question bank results loaded

AP Chemistry Unit 7.11: Ksp and Solubility

Exam points

Covered syllabus points

Covered syllabus points

7.11 Introduction to Solubility Equilibria question 1

Question (a)

Question (i)

Question (ii)