25 mol of formic acid `(HCO_(2)H)` is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The `K_(a)` of formic acid is

Options

A. `6.5xx10^(-3)`
B. `4.3xx10^(-4)`
C. `1.7xx10^(-6)`
D. `5.3xx10^(-2)`

Answer

To find the acid dissociation constant \( K_a \) of formic acid (HCOOH), we can follow these steps:

### Step 1: Calculate the concentration of hydronium ions \([H_3O^+]\)
Given the pH of the solution is 2.19, we can calculate the concentration of hydronium ions using the formula:

\[ [H_3O^+] = 10^{-\text{pH}} = 10^{-2.19} \]

Calculating this gives:

\[ [H_3O^+] \approx 6.45 \times 10^{-3} \, \text{M} \]

### Step 2: Set up the equilibrium expression for the dissociation of formic acid
The dissociation of formic acid in water can be represented as:

\[ \text{HCOOH} \rightleftharpoons \text{HCOO}^- + \text{H}_3O^+ \]

Let \( x \) be the concentration of \( H_3O^+ \) produced at equilibrium, which we found to be \( 6.45 \times 10^{-3} \, \text{M} \).

### Step 3: Write the equilibrium concentrations
Initially, we have:
- \([HCOOH] = 25 \, \text{M}\)
- \([HCOO^-] = 0 \, \text{M}\)
- \([H_3O^+] = 0 \, \text{M}\)

At equilibrium, the concentrations will be:
- \([HCOOH] = 25 - x \approx 25 - 6.45 \times 10^{-3}\)
- \([HCOO^-] = x \approx 6.45 \times 10^{-3}\)
- \([H_3O^+] = x \approx 6.45 \times 10^{-3}\)

### Step 4: Substitute into the \( K_a \) expression
The expression for the acid dissociation constant \( K_a \) is given by:

\[ K_a = \frac{[HCOO^-][H_3O^+]}{[HCOOH]} \]

Substituting the equilibrium concentrations into the equation:

\[ K_a = \frac{(6.45 \times 10^{-3})(6.45 \times 10^{-3})}{25 - 6.45 \times 10^{-3}} \]

Calculating the denominator:

\[ 25 - 6.45 \times 10^{-3} \approx 25 \, \text{M} \quad (\text{since } 6.45 \times 10^{-3} \text{ is negligible compared to } 25) \]

### Step 5: Calculate \( K_a \)
Now substituting the values into the \( K_a \) expression:

\[ K_a = \frac{(6.45 \times 10^{-3})^2}{25} \]

Calculating \( (6.45 \times 10^{-3})^2 \):

\[ (6.45 \times 10^{-3})^2 \approx 4.15 \times 10^{-5} \]

Now substituting this back into the \( K_a \) expression:

\[ K_a = \frac{4.15 \times 10^{-5}}{25} \approx 1.66 \times 10^{-6} \]

### Final Result
Thus, the \( K_a \) of formic acid is approximately:

\[ K_a \approx 1.66 \times 10^{-6} \]