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 given that 25 moles of formic acid \( (HCOOH) \) are dissolved in 1 liter of solution with a pH of 2.19, we can follow these steps:

### Step 1: Calculate the concentration of hydronium ions \([H_3O^+]\)

The pH of the solution is given as 2.19. We can use the formula for pH to find the concentration of hydronium ions:

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

Rearranging this gives:

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

Calculating this:

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

### Step 2: Set up the equilibrium expression

The dissociation of formic acid can be represented as:

\[ HCOOH \rightleftharpoons HCOO^- + H_3O^+ \]

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: Determine the concentrations at equilibrium

Initially, we have:

- \([HCOOH] = 25 \, \text{M}\)
- \([HCOO^-] = 0 \, \text{M}\)
- \([H_3O^+] = 0 \, \text{M}\)

At equilibrium:

- \([HCOOH] = 25 - x \approx 25 - 6.45 \times 10^{-3} \approx 25 \, \text{M}\) (since \( x \) is very small compared to 25)
- \([HCOO^-] = x = 6.45 \times 10^{-3} \, \text{M}\)
- \([H_3O^+] = x = 6.45 \times 10^{-3} \, \text{M}\)

### Step 4: Write the expression for \( K_a \)

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 expression:

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

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

Calculating \( K_a \):

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

Calculating the numerator:

\[ (6.45 \times 10^{-3})^2 = 4.16 \times 10^{-5} \]

Now substituting back into the equation for \( K_a \):

\[ K_a = \frac{4.16 \times 10^{-5}}{25} = 1.664 \times 10^{-6} \]

Rounding this gives:

\[ K_a \approx 1.67 \times 10^{-6} \]

### Final Answer

Thus, the \( K_a \) of formic acid is approximately \( 1.67 \times 10^{-6} \).

---