AP Chemistry Unit 7.12: Common-Ion Solubility

Practice AP Chemistry Unit 7.12 questions on explaining how common ions change solubility and using Ksp or Q to predict precipitation.

Exam points

Use particle, bonding, and equilibrium models to explain solution properties, phase changes, and behavior.
Interpret structural and experimental evidence to connect molecular interactions with observable properties.

Covered syllabus points

7.12.A—Identify the solubility of a salt, and/or the value of Ksp for the salt, based on the concentration of a common…

[Maximum number: 1]

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 7.12 Common-Ion Effect question 1 — AP Chemistry

The student prepares a second saturated solution of $\mathrm{Sr}(\mathrm{OH})_{2}$ in aqueous $0.10 M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}$ instead of water. Will the value of $\left[\mathrm{OH}^{-}\right]$ in the second solution be greater than, less than, or equal to the value in the first solution? Justify your answer. (Assume constant temperature.)

For the correct answer and a valid justification: 1 point
Less than. Because the $\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ solution already contains a common ion, $\mathrm{Sr}^{2+}(a q)$ , the solubility of $\mathrm{Sr}(\mathrm{OH})_{2}$ will be decreased, resulting in a lower value of $\left[\mathrm{OH}^{-}\right]$ .

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Covered syllabus points

7.12 Common-Ion Effect question 1