You dissolve 3.25 g of sodium oxalate (Na2C2O4) in sufficient water to produce a final solution volume of 100.0 mL. Assuming all of the solid dissolves, calculate the solution pH.
The Correct Answer and Explanation is :
To calculate the pH of the solution formed by dissolving sodium oxalate (Na₂C₂O₄) in water, we need to follow several steps.
1. Dissociation of Sodium Oxalate:
Sodium oxalate is a salt, and when it dissolves in water, it dissociates into sodium ions (Na⁺) and oxalate ions (C₂O₄²⁻):
[
\text{Na₂C₂O₄} \rightarrow 2\text{Na}^+ + \text{C₂O₄}^{2-}
]
Sodium ions (Na⁺) do not affect the pH because they are spectator ions, so we focus on the oxalate ions (C₂O₄²⁻), which are the conjugate base of oxalic acid (H₂C₂O₄).
2. Oxalate Ion and Its Hydrolysis:
Oxalate ions can undergo hydrolysis in water to form hydroxide ions (OH⁻), thus affecting the pH of the solution:
[
\text{C₂O₄}^{2-} + \text{H₂O} \rightleftharpoons \text{HC₂O₄}^- + \text{OH}^-
]
The hydrolysis of C₂O₄²⁻ is a weak base reaction. To calculate the pH, we need the base dissociation constant (Kb) for oxalate, which can be derived from the acid dissociation constant (Ka) of oxalic acid. Oxalic acid is a diprotic acid with two dissociation constants. For simplicity, we use the value for the second dissociation (which is the relevant one for the C₂O₄²⁻ ion), and it is:
[
\text{Ka}_2 = 5.4 \times 10^{-5}
]
Using the relationship between Ka and Kb for conjugate acid-base pairs:
[
\text{Kb} = \frac{K_w}{\text{Ka}_2}
]
where ( K_w = 1.0 \times 10^{-14} ) (the ion product of water at 25°C).
[
\text{Kb} = \frac{1.0 \times 10^{-14}}{5.4 \times 10^{-5}} = 1.85 \times 10^{-10}
]
3. Calculate the Concentration of OH⁻:
Now, we calculate the concentration of OH⁻ produced from the hydrolysis of C₂O₄²⁻. First, determine the molarity of C₂O₄²⁻ in the solution. The molar mass of Na₂C₂O₄ is:
[
\text{Molar mass of Na₂C₂O₄} = 2(22.99) + 2(12.01) + 4(16.00) = 134.00 \, \text{g/mol}
]
The moles of Na₂C₂O₄ in 3.25 g is:
[
\text{moles of Na₂C₂O₄} = \frac{3.25 \, \text{g}}{134.00 \, \text{g/mol}} = 0.0243 \, \text{mol}
]
The concentration of C₂O₄²⁻ in the solution is:
[
\text{Concentration of C₂O₄²⁻} = \frac{0.0243 \, \text{mol}}{0.1 \, \text{L}} = 0.243 \, \text{M}
]
Next, we use the Kb expression for hydrolysis:
[
\text{Kb} = \frac{[\text{HC₂O₄}^-][\text{OH}^-]}{[\text{C₂O₄}^{2-}]}
]
Assuming the concentration of OH⁻ formed is ( x ), we have:
[
1.85 \times 10^{-10} = \frac{x^2}{0.243 – x}
]
Since ( x ) is very small compared to 0.243, we approximate:
[
1.85 \times 10^{-10} = \frac{x^2}{0.243}
]
Solving for ( x ) (the concentration of OH⁻):
[
x^2 = (1.85 \times 10^{-10})(0.243)
]
[
x^2 = 4.49 \times 10^{-11}
]
[
x = \sqrt{4.49 \times 10^{-11}} = 6.7 \times 10^{-6} \, \text{M}
]
4. Calculate the pOH and pH:
The concentration of OH⁻ is ( 6.7 \times 10^{-6} \, \text{M} ), so we can calculate the pOH:
[
\text{pOH} = -\log(6.7 \times 10^{-6}) = 5.18
]
Finally, the pH is related to the pOH by:
[
\text{pH} = 14 – \text{pOH} = 14 – 5.18 = 8.82
]
Conclusion:
The pH of the solution is approximately 8.82, indicating that the solution is mildly basic.