Calculate the hydrolysis constant of the reaction `HCO_2^(-) +H_2O Leftrightarrow HCO_2H+OH^(-)` and find the concentrations of `H_3O^+, OH^-, HCO_2^- and HCO_2H` in a solution of 0:15 M `HCO_2` Na. `K_1 (HCOOH) Leftrightarrow 1.8 xx 10^(-4)4`

Answer

To solve the problem, we need to calculate the hydrolysis constant (Kh) for the reaction:

\[ \text{HCO}_2^- + \text{H}_2\text{O} \leftrightarrow \text{HCO}_2\text{H} + \text{OH}^- \]

We are given the dissociation constant (K₁) for formic acid (HCOOH) as \( K_a = 1.8 \times 10^{-4} \) and the ion product of water \( K_w = 1.0 \times 10^{-14} \).

### Step 1: Calculate the Hydrolysis Constant (Kh)

The hydrolysis constant \( K_h \) can be calculated using the relationship:

\[ K_h = \frac{K_w}{K_a} \]

Substituting the given values:

\[ K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \]

### Step 2: Perform the Calculation

Calculating \( K_h \):

\[ K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = \frac{1.0}{1.8} \times 10^{-14 + 4} = \frac{1.0}{1.8} \times 10^{-10} \]

Now, calculating \( \frac{1.0}{1.8} \):

\[ \frac{1.0}{1.8} \approx 0.5556 \]

So,

\[ K_h \approx 0.5556 \times 10^{-10} = 5.56 \times 10^{-11} \]

### Step 3: Find Concentrations in 0.15 M HCO₂Na Solution

Assuming that the initial concentration of HCO₂Na is 0.15 M, we can set up an equilibrium expression. Let \( x \) be the concentration of \( \text{OH}^- \) produced at equilibrium.

At equilibrium:

- \( [\text{HCO}_2^-] = 0.15 - x \)
- \( [\text{HCO}_2\text{H}] = x \)
- \( [\text{OH}^-] = x \)

Using the hydrolysis constant expression:

\[ K_h = \frac{[\text{HCO}_2\text{H}][\text{OH}^-]}{[\text{HCO}_2^-]} \]

Substituting the equilibrium concentrations:

\[ 5.56 \times 10^{-11} = \frac{x \cdot x}{0.15 - x} \]

Assuming \( x \) is small compared to 0.15, we can approximate:

\[ 5.56 \times 10^{-11} = \frac{x^2}{0.15} \]

### Step 4: Solve for x

Rearranging gives:

\[ x^2 = 5.56 \times 10^{-11} \times 0.15 \]

Calculating:

\[ x^2 = 8.34 \times 10^{-12} \]

Taking the square root:

\[ x = \sqrt{8.34 \times 10^{-12}} \approx 2.89 \times 10^{-6} \]

### Step 5: Calculate Concentrations

Now we can find the concentrations:

- \( [\text{OH}^-] = x \approx 2.89 \times 10^{-6} \)
- \( [\text{HCO}_2\text{H}] = x \approx 2.89 \times 10^{-6} \)
- \( [\text{HCO}_2^-] = 0.15 - x \approx 0.15 \) (since \( x \) is very small)
- To find \( [\text{H}_3\text{O}^+] \), we can use the relation \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \):

\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{2.89 \times 10^{-6}} \approx 3.47 \times 10^{-9} \]

### Final Concentrations

- \( [\text{H}_3\text{O}^+] \approx 3.47 \times 10^{-9} \, \text{M} \)
- \( [\text{OH}^-] \approx 2.89 \times 10^{-6} \, \text{M} \)
- \( [\text{HCO}_2^-] \approx 0.15 \, \text{M} \)
- \( [\text{HCO}_2\text{H}] \approx 2.89 \times 10^{-6} \, \text{M} \)