Determine K for the reaction :
`H_2C_2O_4(aq)+2OH^(-)(aq) to C_2O_4^(2-)(aq)+2H_2O(l)`
`H_2C_2O_4(aq)K_(a_1)=6.5xx10^(-2), K_(a_2)=6.1xx10^(-5)`
`H_2O K_w=1.0xx10^(-14)`

Options

A. `4.0xx10^(-34)`
B. `4.0xx10^(-6)`
C. `4.0xx10^(6)`
D. `4.0xx10^(22)`

Answer

To determine the equilibrium constant \( K \) for the reaction

\[ H_2C_2O_4(aq) + 2OH^-(aq) \rightarrow C_2O_4^{2-}(aq) + 2H_2O(l) \]

we will use the provided dissociation constants \( K_{a_1} \), \( K_{a_2} \), and \( K_w \).

### Step 1: Write the dissociation reactions and their constants

1. The first dissociation of oxalic acid \( H_2C_2O_4 \):
\[ H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^- \quad K_{a_1} = 6.5 \times 10^{-2} \]

2. The second dissociation of oxalic acid:
\[ HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-} \quad K_{a_2} = 6.1 \times 10^{-5} \]

3. The ion product of water:
\[ H_2O \rightleftharpoons H^+ + OH^- \quad K_w = 1.0 \times 10^{-14} \]

### Step 2: Combine the reactions to find the desired equilibrium constant

To find the equilibrium constant for the overall reaction, we can use the following approach:

- The reaction we want can be derived from the first and second dissociation reactions of oxalic acid and the ionization of water.

We can express the desired reaction as:

\[ H_2C_2O_4 + 2OH^- \rightarrow C_2O_4^{2-} + 2H_2O \]

This can be derived by combining the first dissociation and the second dissociation, and then subtracting twice the ionization of water.

### Step 3: Write the equilibrium constant expression

The equilibrium constant \( K \) for the overall reaction can be expressed as:

\[ K = \frac{K_{a_1} \cdot K_{a_2}}{K_w^2} \]

### Step 4: Substitute the values of \( K_{a_1} \), \( K_{a_2} \), and \( K_w \)

Substituting the known values:

\[ K = \frac{(6.5 \times 10^{-2}) \cdot (6.1 \times 10^{-5})}{(1.0 \times 10^{-14})^2} \]

### Step 5: Calculate \( K \)

1. Calculate the numerator:
\[ 6.5 \times 10^{-2} \cdot 6.1 \times 10^{-5} = 3.965 \times 10^{-6} \]

2. Calculate the denominator:
\[ (1.0 \times 10^{-14})^2 = 1.0 \times 10^{-28} \]

3. Now calculate \( K \):
\[ K = \frac{3.965 \times 10^{-6}}{1.0 \times 10^{-28}} = 3.965 \times 10^{22} \]

4. Rounding gives:
\[ K \approx 4.0 \times 10^{22} \]

### Final Result

Thus, the equilibrium constant \( K \) for the reaction is:

\[ K \approx 4.0 \times 10^{22} \]