Calcium sulfate dihydrate (CaSO4 • 2H2O) has a molar mass of 172.17 g/mol. Considering that calcium sulfate dihydrate is a salt, what would be the pH of a 200 mL aqueous solution containing 1.45 g of the dissolved salt? The Ka2 value for sulfuric acid is 1.2 × 10–2.
The Correct Answer and Explanation is:
To find the pH of the 200 mL aqueous solution containing 1.45 g of calcium sulfate dihydrate (CaSO₄ · 2H₂O), we need to follow these steps:
Step 1: Determine moles of CaSO₄ · 2H₂O
Given:
- Mass = 1.45 g
- Molar mass = 172.17 g/mol
Moles=1.45 g172.17 g/mol≈0.00842 mol\text{Moles} = \frac{1.45\ \text{g}}{172.17\ \text{g/mol}} \approx 0.00842\ \text{mol}Moles=172.17 g/mol1.45 g≈0.00842 mol
Step 2: Find concentration in 200 mL solution
Volume=200 mL=0.200 L\text{Volume} = 200\ \text{mL} = 0.200\ \text{L}Volume=200 mL=0.200 LConcentration=0.00842 mol0.200 L=0.0421 M\text{Concentration} = \frac{0.00842\ \text{mol}}{0.200\ \text{L}} = 0.0421\ \text{M}Concentration=0.200 L0.00842 mol=0.0421 M
Step 3: Dissociation in water
Calcium sulfate dissolves as:CaSO₄→Ca2++SO₄2−\text{CaSO₄} \rightarrow \text{Ca}^{2+} + \text{SO₄}^{2-}CaSO₄→Ca2++SO₄2−
The sulfate ion (SO₄²⁻) is the conjugate base of the hydrogen sulfate ion (HSO₄⁻), which has a Ka₂ = 1.2 × 10⁻².
We use this Ka₂ to assess the basicity of SO₄²⁻ via its Kb:Kw=1.0×10−14(at 25°C)K_w = 1.0 \times 10^{-14} \quad \text{(at 25°C)}Kw=1.0×10−14(at 25°C)Kb=KwKa=1.0×10−141.2×10−2=8.33×10−13K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-2}} = 8.33 \times 10^{-13}Kb=KaKw=1.2×10−21.0×10−14=8.33×10−13
Step 4: Use ICE table for SO₄²⁻ hydrolysis
SO₄2−+H2O↔HSO4−+OH−\text{SO₄}^{2-} + H₂O \leftrightarrow HSO₄⁻ + OH⁻SO₄2−+H2O↔HSO4−+OH−
Let x be the concentration of OH⁻ formed:Kb=x20.0421=8.33×10−13K_b = \frac{x^2}{0.0421} = 8.33 \times 10^{-13}Kb=0.0421×2=8.33×10−13×2=(8.33×10−13)(0.0421)=3.51×10−14x^2 = (8.33 \times 10^{-13})(0.0421) = 3.51 \times 10^{-14}x2=(8.33×10−13)(0.0421)=3.51×10−14x=3.51×10−14≈5.93×10−8 M OH⁻x = \sqrt{3.51 \times 10^{-14}} \approx 5.93 \times 10^{-8}\ \text{M OH⁻}x=3.51×10−14≈5.93×10−8 M OH⁻
Step 5: Calculate pOH and then pH
pOH=−log[OH−]=−log(5.93×10−8)≈7.23\text{pOH} = -\log[OH⁻] = -\log(5.93 \times 10^{-8}) \approx 7.23pOH=−log[OH−]=−log(5.93×10−8)≈7.23pH=14−7.23=6.77\text{pH} = 14 – 7.23 = 6.77pH=14−7.23=6.77
Final Answer:
The pH is approximately 6.77
Explanation
Calcium sulfate dihydrate (CaSO₄ · 2H₂O) is a neutral salt derived from a strong base (Ca(OH)₂) and a diprotic strong acid (H₂SO₄). The second dissociation step of sulfuric acid is weak, and it determines the acidic or basic character of the sulfate ion in water. To calculate the pH, we start by finding how many moles of salt are present. Given a mass of 1.45 g and a molar mass of 172.17 g/mol, this gives about 0.00842 mol in 200 mL of water, resulting in a 0.0421 M solution.
When dissolved, calcium sulfate releases Ca²⁺ and SO₄²⁻ ions. The calcium ion does not affect pH. The sulfate ion can react with water slightly, forming some hydroxide ions. Since sulfate is the conjugate base of HSO₄⁻, we calculate its basicity using the given Ka₂ of sulfuric acid. Using the relationship Kb = Kw / Ka, we find that Kb for SO₄²⁻ is 8.33 × 10⁻¹³.
We then solve for the hydroxide ion concentration using an ICE table and the equation for Kb. This gives an OH⁻ concentration of approximately 5.93 × 10⁻⁸ M. The pOH is calculated from this value and subtracted from 14 to find the pH.
The final result, a pH of 6.77, indicates that the solution is very slightly acidic but very close to neutral. This makes sense for a salt formed from a strong base and a weak acid, where the conjugate base only weakly hydrolyzes in water.
