4) Find the [H3O*] of a 0.100 M HCN solution. Ka for HCN is 4.9 x 10-10

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**Problem Statement:**

Find the [H₃O⁺] of a 0.100 M HCN solution. The Ka for HCN is 4.9 × 10⁻¹⁰.

---

**Explanation:**

This problem involves calculating the hydronium ion concentration [H₃O⁺] of a hydrocyanic acid (HCN) solution using its dissociation constant (Ka).

1. **Write the dissociation equation for HCN:**

   \[
   \text{HCN} \leftrightarrow \text{H⁺} + \text{CN⁻}
   \]

   In water, H⁺ is equivalent to H₃O⁺, so the equation can be written as:

   \[
   \text{HCN} + \text{H₂O} \leftrightarrow \text{H₃O⁺} + \text{CN⁻}
   \]

2. **Write the expression for Ka:**

   \[
   \text{Ka} = \frac{[\text{H₃O⁺}][\text{CN⁻}]}{[\text{HCN}]}
   \]

3. **Assume initial concentrations and change:**

   - Initial [HCN] = 0.100 M
   - Initial [H₃O⁺] = 0
   - Change: [HCN] decreases by x, [H₃O⁺] and [CN⁻] increase by x

   Therefore:

   - [HCN] = 0.100 - x
   - [H₃O⁺] = x
   - [CN⁻] = x

4. **Substitute into the Ka expression:**

   \[
   \text{Ka} = \frac{x \times x}{0.100 - x} = \frac{x^2}{0.100 - x}
   \]

   Assume x is small compared to 0.100 (since HCN is a weak acid), then:

   \[
   \text{Ka} \approx \frac{x^2}{0.100}
   \]

5. **Solve for x:**

   \[
   4.9 \times 10^{-10} = \frac{x^2}{0.100}
   \]

   \
Transcribed Image Text:**Problem Statement:** Find the [H₃O⁺] of a 0.100 M HCN solution. The Ka for HCN is 4.9 × 10⁻¹⁰. --- **Explanation:** This problem involves calculating the hydronium ion concentration [H₃O⁺] of a hydrocyanic acid (HCN) solution using its dissociation constant (Ka). 1. **Write the dissociation equation for HCN:** \[ \text{HCN} \leftrightarrow \text{H⁺} + \text{CN⁻} \] In water, H⁺ is equivalent to H₃O⁺, so the equation can be written as: \[ \text{HCN} + \text{H₂O} \leftrightarrow \text{H₃O⁺} + \text{CN⁻} \] 2. **Write the expression for Ka:** \[ \text{Ka} = \frac{[\text{H₃O⁺}][\text{CN⁻}]}{[\text{HCN}]} \] 3. **Assume initial concentrations and change:** - Initial [HCN] = 0.100 M - Initial [H₃O⁺] = 0 - Change: [HCN] decreases by x, [H₃O⁺] and [CN⁻] increase by x Therefore: - [HCN] = 0.100 - x - [H₃O⁺] = x - [CN⁻] = x 4. **Substitute into the Ka expression:** \[ \text{Ka} = \frac{x \times x}{0.100 - x} = \frac{x^2}{0.100 - x} \] Assume x is small compared to 0.100 (since HCN is a weak acid), then: \[ \text{Ka} \approx \frac{x^2}{0.100} \] 5. **Solve for x:** \[ 4.9 \times 10^{-10} = \frac{x^2}{0.100} \] \
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