Calculate `[H^+],[HCOO^(-)] and [OCN^(-)]` in a solution that contains 0.1 M HCOOH `(K_a=2.4xx10^(-4)) and 0.1 M HOCN (K_a=4xx10^(-4))`.

Options

A. `[H^+]=8xx10^(-3)M`
B. `[HCOO^(-)]=3xx10^(-3)M`
C. `[OCN^(-)]=5xx10^(-3)M`
D. All of these

Explanation

To solve the problem of calculating the concentrations of \([H^+]\), \([HCOO^-]\), and \([OCN^-]\) in a solution containing 0.1 M formic acid (HCOOH) and 0.1 M cyanic acid (HOCN), we can follow these steps:

### Step 1: Calculate \([H^+]\) from HCOOH

1. **Write the dissociation equation for HCOOH**:
\[ HCOOH \rightleftharpoons H^+ + HCOO^- \]

2. **Set up the expression for \(K_a\)**:
\[ K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} \]
Given \(K_a = 2.4 \times 10^{-4}\) and initial concentration of HCOOH is 0.1 M.

3. **Assume \(x\) is the concentration of \(H^+\) produced**:
\[ K_a = \frac{x \cdot x}{0.1 - x} \approx \frac{x^2}{0.1} \quad (\text{assuming } x \text{ is small compared to } 0.1) \]

4. **Substitute \(K_a\) into the equation**:
\[ 2.4 \times 10^{-4} = \frac{x^2}{0.1} \]

5. **Solve for \(x\)**:
\[ x^2 = 2.4 \times 10^{-4} \times 0.1 = 2.4 \times 10^{-5} \]
\[ x = \sqrt{2.4 \times 10^{-5}} \approx 4.9 \times 10^{-3} \text{ M} \]

### Step 2: Calculate \([HCOO^-]\)

1. **From the dissociation of HCOOH, we know**:
\[ [HCOO^-] = x \approx 4.9 \times 10^{-3} \text{ M} \]

### Step 3: Calculate \([OCN^-]\) from HOCN

1. **Write the dissociation equation for HOCN**:
\[ HOCN \rightleftharpoons H^+ + OCN^- \]

2. **Set up the expression for \(K_a\)**:
\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]
Given \(K_a = 4 \times 10^{-4}\) and initial concentration of HOCN is 0.1 M.

3. **Assume \(y\) is the concentration of \(OCN^-\) produced**:
\[ K_a = \frac{(4.9 \times 10^{-3}) \cdot y}{0.1 - y} \approx \frac{(4.9 \times 10^{-3}) \cdot y}{0.1} \]

4. **Substitute \(K_a\) into the equation**:
\[ 4 \times 10^{-4} = \frac{(4.9 \times 10^{-3}) \cdot y}{0.1} \]

5. **Solve for \(y\)**:
\[ y = \frac{4 \times 10^{-4} \times 0.1}{4.9 \times 10^{-3}} \approx 8.16 \times 10^{-3} \text{ M} \]

### Final Results

- \([H^+] \approx 4.9 \times 10^{-3} \text{ M}\)
- \([HCOO^-] \approx 4.9 \times 10^{-3} \text{ M}\)
- \([OCN^-] \approx 8.16 \times 10^{-3} \text{ M}\)