`K_(1)` & `K_(2)` for oxalic acid are `6.5 xx 10^(-2)` and `6.1 xx 10^(-5)` respectively . What will be the `[OH^(-)]` in a `0.01 M` solution of sodium oxalate

Options

A. `9.6xx10^(-6)`
B. `1.4xx10^(-6)`
C. `1.28xx10^(-6)`
D. `1.3xx10^(-8)`

Answer

To find the concentration of hydroxide ions \([OH^-]\) in a \(0.01 \, M\) solution of sodium oxalate, we will follow these steps:

### Step 1: Understand the dissociation of sodium oxalate
Sodium oxalate (\(Na_2C_2O_4\)) dissociates in water to give sodium ions and oxalate ions:
\[ Na_2C_2O_4 \rightarrow 2Na^+ + C_2O_4^{2-} \]
Since the concentration of sodium oxalate is \(0.01 \, M\), the concentration of oxalate ions \([C_2O_4^{2-}]\) will also be \(0.01 \, M\).

### Step 2: Hydrolysis of the oxalate ion
The oxalate ion can undergo hydrolysis to produce hydroxide ions:
\[ C_2O_4^{2-} + H_2O \rightleftharpoons HC_2O_4^- + OH^- \]

### Step 3: Set up the equilibrium expression
Let \(x\) be the concentration of hydroxide ions produced at equilibrium. The equilibrium concentrations will be:
- \([C_2O_4^{2-}] = 0.01 - x\)
- \([HC_2O_4^-] = x\)
- \([OH^-] = x\)

### Step 4: Write the hydrolysis constant expression
The hydrolysis constant \(K_h\) can be expressed as:
\[ K_h = \frac{[HC_2O_4^-][OH^-]}{[C_2O_4^{2-}]} \]
Substituting the equilibrium concentrations:
\[ K_h = \frac{x \cdot x}{0.01 - x} = \frac{x^2}{0.01 - x} \]

### Step 5: Relate \(K_h\) to \(K_w\) and \(K_2\)
The hydrolysis constant \(K_h\) can also be expressed in terms of the dissociation constants of oxalic acid:
\[ K_h = \frac{K_w}{K_2} \]
Where \(K_w = 1.0 \times 10^{-14}\) and \(K_2 = 6.1 \times 10^{-5}\).

### Step 6: Calculate \(K_h\)
Substituting the values:
\[ K_h = \frac{1.0 \times 10^{-14}}{6.1 \times 10^{-5}} \approx 1.64 \times 10^{-10} \]

### Step 7: Set the equation for \(x\)
Now, we can set the equation:
\[ 1.64 \times 10^{-10} = \frac{x^2}{0.01} \]
Rearranging gives:
\[ x^2 = 1.64 \times 10^{-10} \times 0.01 = 1.64 \times 10^{-12} \]

### Step 8: Solve for \(x\)
Taking the square root:
\[ x = \sqrt{1.64 \times 10^{-12}} \approx 1.28 \times 10^{-6} \, M \]

### Conclusion
Thus, the concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] = 1.28 \times 10^{-6} \, M \]