Multiple Choice
Which term describes the ability of a substance to dissolve in another substance?
3
views
6. Chemical Quantities & Aqueous Reactions
Solubility Rules
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Given that the solubility product constant (K_sp) for Ca(OH)_2 at 25°C is 5.5 × 10^{-6}, what is the molar solubility of Ca(OH)_2 in pure water?
A
1.2 × 10^{-2} mol/L
B
1.2 × 10^{-2} mol/L
C
1.1 × 10^{-2} mol/L
D
1.1 × 10^{-2} mol/L
Verified step by step guidance1
Write the dissociation equation for calcium hydroxide: $\mathrm{Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 OH^- (aq)}$.
Define the molar solubility of $\mathrm{Ca(OH)_2}$ as $s$, which means the concentration of $\mathrm{Ca^{2+}}$ ions is $s$ and the concentration of $\mathrm{OH^-}$ ions is $2s$ because each formula unit produces two hydroxide ions.
Write the expression for the solubility product constant $K_{sp}$ in terms of $s$: $K_{sp} = [Ca^{2+}][OH^-]^2 = s \times (2s)^2 = 4s^3$.
Set the expression equal to the given $K_{sp}$ value and solve for $s$: $4s^3 = 5.5 \times 10^{-6}$.
Rearrange to isolate $s$: $s^3 = \frac{5.5 \times 10^{-6}}{4}$, then take the cube root of both sides to find the molar solubility $s$.
2:07mWatch next
Master Solubility and Insolubility with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
Solubility Rules practice set
Problem sets built by lead tutorsExpert video explanations