Chemistry

Chemistry

Problem

At 25 °C, the solubility‐product constant for CaF₂ is Ksp=1.5×10−10K_{sp}=1.5\times10^{-10}. Determine the percent decrease in the molar solubility of CaF₂ when it is added to a 0.0050 M NaF solution rather than pure water. Give your answer to the nearest tenth of a percent.
Intuitive explanation

The key here is the common ion effect: when you add a soluble source of one of the ions already in the equilibrium (here, F−\text{F}^- from NaF), it shifts the dissolution equilibrium backward, reducing how much CaF₂ can dissolve. Set up the KspK_{sp} expression for both scenarios—pure water and with excess F−\text{F}^- present—and compare the resulting molar solubilities. The F−\text{F}^- concentration from NaF dominates the equilibrium, so you can use it as a known quantity in your second calculation.

Intuitive explanation

The key here is the common ion effect: when you add a soluble source of one of the ions already in the equilibrium (here, F−\text{F}^- from NaF), it shifts the dissolution equilibrium backward, reducing how much CaF₂ can dissolve. Set up the KspK_{sp} expression for both scenarios—pure water and with excess F−\text{F}^- present—and compare the resulting molar solubilities. The F−\text{F}^- concentration from NaF dominates the equilibrium, so you can use it as a known quantity in your second calculation.

Browse the problem library