The compound lead(II) chromate (PbCrO4, molar mass = 323.194 g/mol) is only slightly soluble in water. Its Ksp at 25°C is 1.8×10-14. Estimate the solubility of lead(II) chromate in water in units of grams per 100 mL of water. g per 100 mL
The Correct Answer and Explanation is:
To estimate the solubility of lead(II) chromate (PbCrO₄) in water in grams per 100 mL, we first determine its solubility in mol/L (molar solubility) using its Ksp value, then convert that to grams per 100 mL.
Step 1: Dissociation Equation
Lead(II) chromate dissociates in water as: PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)\text{PbCrO}_4 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{CrO}_4^{2-} (aq)PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)
Let the solubility be s mol/L, then:
- [Pb2+]=s[Pb^{2+}] = s[Pb2+]=s
- [CrO42−]=s[CrO_4^{2-}] = s[CrO42−]=s
Step 2: Ksp Expression
Ksp=[Pb2+][CrO42−]=s⋅s=s2K_{sp} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}] = s \cdot s = s^2Ksp=[Pb2+][CrO42−]=s⋅s=s2 1.8×10−14=s21.8 \times 10^{-14} = s^21.8×10−14=s2 s=1.8×10−14=1.34×10−7 mol/Ls = \sqrt{1.8 \times 10^{-14}} = 1.34 \times 10^{-7} \text{ mol/L}s=1.8×10−14=1.34×10−7 mol/L
Step 3: Convert molar solubility to grams per liter
Molar mass of PbCrO4=323.194 g/mol\text{Molar mass of PbCrO}_4 = 323.194 \text{ g/mol}Molar mass of PbCrO4=323.194 g/mol Solubility (g/L)=1.34×10−7 mol/L×323.194 g/mol≈4.33×10−5 g/L\text{Solubility (g/L)} = 1.34 \times 10^{-7} \text{ mol/L} \times 323.194 \text{ g/mol} \approx 4.33 \times 10^{-5} \text{ g/L}Solubility (g/L)=1.34×10−7 mol/L×323.194 g/mol≈4.33×10−5 g/L
Step 4: Convert g/L to g per 100 mL
Solubility (g/100 mL)=4.33×10−5 g1000 mL×100 mL=4.33×10−6 g/100 mL\text{Solubility (g/100 mL)} = \frac{4.33 \times 10^{-5} \text{ g}}{1000 \text{ mL}} \times 100 \text{ mL} = 4.33 \times 10^{-6} \text{ g/100 mL}Solubility (g/100 mL)=1000 mL4.33×10−5 g×100 mL=4.33×10−6 g/100 mL
✅ Final Answer:
4.33×10−6 g per 100 mL\boxed{4.33 \times 10^{-6} \text{ g per 100 mL}}4.33×10−6 g per 100 mL
🧪 Explanation (300+ words):
Solubility is a measure of how much of a substance can dissolve in a given amount of solvent. For sparingly soluble salts like lead(II) chromate (PbCrO₄), solubility is often calculated using the solubility product constant (Ksp). The Ksp is an equilibrium constant that reflects the extent to which a salt dissociates into its ions in solution.
When PbCrO₄ dissolves, it breaks into Pb²⁺ and CrO₄²⁻ ions in a 1:1 molar ratio. If we let “s” represent the molar solubility (in mol/L), then the equilibrium concentrations of both Pb²⁺ and CrO₄²⁻ will be “s.” The Ksp expression for this equilibrium is then Ksp=[Pb2+][CrO42−]=s2K_{sp} = [Pb^{2+}][CrO_4^{2-}] = s^2Ksp=[Pb2+][CrO42−]=s2. By substituting the known Ksp value of 1.8×10−141.8 \times 10^{-14}1.8×10−14, we can solve for “s,” the molar solubility.
Once the molar solubility is determined (approximately 1.34×10−71.34 \times 10^{-7}1.34×10−7 mol/L), we convert it to mass by multiplying it by the molar mass of PbCrO₄ (323.194 g/mol), resulting in about 4.33×10−54.33 \times 10^{-5}4.33×10−5 g/L. Since the question asks for grams per 100 mL, we scale this down by a factor of 10 (because 100 mL is one-tenth of a liter), yielding 4.33×10−64.33 \times 10^{-6}4.33×10−6 g per 100 mL.
This very low solubility reflects the fact that PbCrO₄ is highly insoluble in water—a property common among heavy metal chromates. Such data is important in environmental chemistry, where understanding the mobility and bioavailability of heavy metal compounds is critical for pollution control and water safety.