Question

Slaked lime $\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]$ is used to soften hard water by removing calcium ions from hard water through the reaction $$\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow{2 \mathrm{CaCO}_{3}(s)}+2 \mathrm{H}_{2} \mathrm{O}(l)$$ Although $\mathrm{CaCO}_{3}(s)$ is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of $\mathrm{CaCO}_{3}$ in water $\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right)$

Name: Problem 20, Chapter 18: Chemical Principle
Uploaded: 2020-08-10T03:28:30-08:00
Duration: 2 min 35 s
Channel: Bin Chen
Description: Slaked lime [Ca(OH)2] is used to soften hard water by removing calcium ions from hard water through the reaction Ca(OH)2(a q)+Ca^2+(a q)+2 HCO3^-(a q) ⟶2 CaCO3(s)+2 H2 O(l) Although CaCO3(s) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of CaCO3 in water (Ksp=8.7 ×10^-9)

   Slaked lime $\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]$ is used to soften hard water by removing calcium ions from hard water through the reaction
$$\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow{2 \mathrm{CaCO}_{3}(s)}+2 \mathrm{H}_{2} \mathrm{O}(l)$$
Although $\mathrm{CaCO}_{3}(s)$ is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of $\mathrm{CaCO}_{3}$ in water $\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right)$

Chemical Principle

Steven S. Zumdahl,… 7th Edition

Chapter 18, Problem 20

Instant Answer

Solved by Expert Bin Chen

08/10/2020

Step 1

The equation is: $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$$ Show more…

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Slaked lime $\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]$ is used to soften hard water by removing calcium ions from hard water through the reaction $$\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow{2 \mathrm{CaCO}_{3}(s)}+2 \mathrm{H}_{2} \mathrm{O}(l)$$ Although $\mathrm{CaCO}_{3}(s)$ is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of $\mathrm{CaCO}_{3}$ in water $\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right)$

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Key Concepts

Solubility Product Constant (Ksp)

This constant represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound. It quantifies the maximum product of the ion concentrations (each raised to the power of its stoichiometric coefficient) that can exist in solution before the formation of a precipitate.

Dissolution Equilibrium

Dissolution equilibrium is the dynamic balance between a sparingly soluble solid and its constituent ions in solution. At equilibrium, the rate of dissolution equals the rate of precipitation, and the ion concentrations remain constant, governed by the solubility product constant (Ksp).

Molar Solubility

Molar solubility is the measure of the number of moles of a solute that can dissolve in one liter of solution to form a saturated solution. It can be calculated by setting up the equilibrium expression for the dissolution of the compound and solving for the concentration of the dissolved ions using the Ksp value.

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