The Ksp for BaSO4 is 1.1×10^-10 at 25 °C. Calculate the solubility of barium sulfate in pure water in moles per liter and grams per liter.
The Correct Answer and Explanation is:
The solubility product (Ksp) of barium sulfate, BaSO4, at 25°C is given as 1.1 x 10^-10. We can use this value to calculate the solubility of BaSO4 in pure water in both moles per liter and grams per liter.
Step 1: Write the dissociation equation for BaSO4
When BaSO4 dissolves in water, it dissociates according to the equation: BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq)BaSO4(s)⇌Ba2+(aq)+SO42−(aq)
The stoichiometry of the dissociation tells us that for every mole of BaSO4 that dissolves, one mole of Ba²⁺ and one mole of SO₄²⁻ are produced.
Step 2: Set up the expression for Ksp
The solubility product expression for BaSO4 is: Ksp=[Ba2+][SO42−]K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]Ksp=[Ba2+][SO42−]
Let the solubility of BaSO4 be s moles per liter. At equilibrium: [Ba2+]=sand[SO42−]=s[\text{Ba}^{2+}] = s \quad \text{and} \quad [\text{SO}_4^{2-}] = s[Ba2+]=sand[SO42−]=s
Therefore, the Ksp expression becomes: Ksp=s2K_{sp} = s^2Ksp=s2
Step 3: Solve for s
Given that Ksp=1.1×10−10K_{sp} = 1.1 \times 10^{-10}Ksp=1.1×10−10, we can substitute this value into the equation: 1.1×10−10=s21.1 \times 10^{-10} = s^21.1×10−10=s2
Solving for s: s=1.1×10−10=1.048×10−5 mol/Ls = \sqrt{1.1 \times 10^{-10}} = 1.048 \times 10^{-5} \, \text{mol/L}s=1.1×10−10=1.048×10−5mol/L
So, the solubility of BaSO4 in pure water is approximately 1.05×10−5 mol/L1.05 \times 10^{-5} \, \text{mol/L}1.05×10−5mol/L.
Step 4: Convert the solubility to grams per liter
Now, to convert the solubility from moles per liter to grams per liter, we use the molar mass of BaSO4. The molar mass is calculated as: Molar mass of BaSO4=137.33 (Ba)+32.07 (S)+4×16.00 (O)=233.39 g/mol\text{Molar mass of BaSO}_4 = 137.33 \, (\text{Ba}) + 32.07 \, (\text{S}) + 4 \times 16.00 \, (\text{O}) = 233.39 \, \text{g/mol}Molar mass of BaSO4=137.33(Ba)+32.07(S)+4×16.00(O)=233.39g/mol
Now, we can calculate the solubility in grams per liter: Solubility (g/L)=s×Molar mass=(1.05×10−5 mol/L)×233.39 g/mol=2.45×10−3 g/L\text{Solubility (g/L)} = s \times \text{Molar mass} = (1.05 \times 10^{-5} \, \text{mol/L}) \times 233.39 \, \text{g/mol} = 2.45 \times 10^{-3} \, \text{g/L}Solubility (g/L)=s×Molar mass=(1.05×10−5mol/L)×233.39g/mol=2.45×10−3g/L
Final Answer:
- Solubility in moles per liter: 1.05×10−5 mol/L1.05 \times 10^{-5} \, \text{mol/L}1.05×10−5mol/L
- Solubility in grams per liter: 2.45×10−3 g/L2.45 \times 10^{-3} \, \text{g/L}2.45×10−3g/L
This calculation shows how to determine both the molar and mass solubility of BaSO4 using its solubility product (Ksp).
