What is the solubility of copper hydroxide Cu(OH)2 in mL? Mwt of Cu(OH)2 = 97.57; Ksp = 5.5 x 10^21, Cu^2+ (aq) + 2 OH^- → Cu(OH)2 (s) (aq)
The Correct Answer and Explanation is:
To determine the solubility of copper hydroxide (Cu(OH)2) in mL, we first need to use the given solubility product constant (Ksp) and molar mass of Cu(OH)2 to find its solubility in moles per liter (mol/L) and then convert it into mL.
Step 1: Write the dissociation equation for Cu(OH)2
The dissociation of Cu(OH)2 in water is given by:
Cu(OH)2(s)⇌Cu2+(aq)+2OH−(aq)\text{Cu(OH)}_2 (s) \rightleftharpoons \text{Cu}^{2+} (aq) + 2 \text{OH}^- (aq)Cu(OH)2(s)⇌Cu2+(aq)+2OH−(aq)
The solubility product constant (Ksp) expression for this dissociation is:
Ksp=[Cu2+][OH−]2K_{sp} = [\text{Cu}^{2+}] [\text{OH}^-]^2Ksp=[Cu2+][OH−]2
Step 2: Define solubility in terms of molarity
Let the solubility of Cu(OH)2 in mol/L be SSS. When Cu(OH)2 dissolves, for every 1 mole of Cu(OH)2 that dissolves, 1 mole of Cu²⁺ and 2 moles of OH⁻ are produced. Therefore:
- [Cu²⁺] = SSS (in mol/L)
- [OH⁻] = 2S (in mol/L)
Step 3: Substitute values into the Ksp equation
Substitute the expressions for [Cu²⁺] and [OH⁻] into the Ksp equation:
Ksp=(S)(2S)2=4S3K_{sp} = (S)(2S)^2 = 4S^3Ksp=(S)(2S)2=4S3
Given Ksp=5.5×10−21K_{sp} = 5.5 \times 10^{-21}Ksp=5.5×10−21, we can solve for SSS:5.5×10−21=4S35.5 \times 10^{-21} = 4S^35.5×10−21=4S3S3=5.5×10−214S^3 = \frac{5.5 \times 10^{-21}}{4}S3=45.5×10−21S3=1.375×10−21S^3 = 1.375 \times 10^{-21}S3=1.375×10−21S=1.375×10−213≈1.12×10−7 mol/LS = \sqrt[3]{1.375 \times 10^{-21}} \approx 1.12 \times 10^{-7} \, \text{mol/L}S=31.375×10−21≈1.12×10−7mol/L
Step 4: Convert solubility to grams per liter
To find the solubility in grams per liter, multiply the solubility in moles per liter by the molar mass of Cu(OH)2:Solubility (g/L)=S×Molar Mass=1.12×10−7 mol/L×97.57 g/mol\text{Solubility (g/L)} = S \times \text{Molar Mass} = 1.12 \times 10^{-7} \, \text{mol/L} \times 97.57 \, \text{g/mol}Solubility (g/L)=S×Molar Mass=1.12×10−7mol/L×97.57g/molSolubility (g/L)≈1.09×10−5 g/L\text{Solubility (g/L)} \approx 1.09 \times 10^{-5} \, \text{g/L}Solubility (g/L)≈1.09×10−5g/L
Step 5: Convert to mL
To express the solubility in mL, we note that 1 L = 1000 mL. Therefore, the solubility in mL is:Solubility (mL)=1.09×10−5 g/L×1000 mL/L=1.09×10−2 mL/L\text{Solubility (mL)} = 1.09 \times 10^{-5} \, \text{g/L} \times 1000 \, \text{mL/L} = 1.09 \times 10^{-2} \, \text{mL/L}Solubility (mL)=1.09×10−5g/L×1000mL/L=1.09×10−2mL/L
Thus, the solubility of Cu(OH)2 is approximately 1.09 × 10^-2 mL/L, which reflects a very low solubility.
