Hydrocyanic acid has a Ka value of 4.9 x 10-10. If the initial concentration of HCN is 0.432 M determine the pOH of this solution at equilibrium.

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### Determining the pOH of a Hydrocyanic Acid Solution

**Problem Statement:**

Hydrocyanic acid (HCN) has a \( K_a \) value of \( 4.9 \times 10^{-10} \). If the initial concentration of HCN is 0.432 M, determine the pOH of this solution at equilibrium.

---

**Solution Steps:**

1. **Write the ionization equation for HCN:**
   
   \[
   \text{HCN} \, \rightleftharpoons \, \text{H}^+ + \text{CN}^-
   \]

2. **Set up the expression for the acid dissociation constant \( K_a \):**

   \[
   K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}
   \]

   Given \( K_a = 4.9 \times 10^{-10} \) and initial [HCN] = 0.432 M.

3. **Assume \( x \) is the concentration of \( \text{H}^+ \) and \( \text{CN}^- \) at equilibrium:**

   \[
   K_a = \frac{x^2}{0.432 - x} \approx \frac{x^2}{0.432}
   \]

4. **Solve for \( x \):**

   \[
   x^2 = 4.9 \times 10^{-10} \times 0.432
   \]

   Calculate \( x \), which represents \( [\text{H}^+] \).

5. **Calculate the pH:**

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

6. **Find pOH using the relation:**

   \[
   \text{pOH} = 14 - \text{pH}
   \]

This step-by-step guide helps in determining the equilibrium pOH of a weak acid solution using its dissociation constant and initial concentration.
Transcribed Image Text:### Determining the pOH of a Hydrocyanic Acid Solution **Problem Statement:** Hydrocyanic acid (HCN) has a \( K_a \) value of \( 4.9 \times 10^{-10} \). If the initial concentration of HCN is 0.432 M, determine the pOH of this solution at equilibrium. --- **Solution Steps:** 1. **Write the ionization equation for HCN:** \[ \text{HCN} \, \rightleftharpoons \, \text{H}^+ + \text{CN}^- \] 2. **Set up the expression for the acid dissociation constant \( K_a \):** \[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \] Given \( K_a = 4.9 \times 10^{-10} \) and initial [HCN] = 0.432 M. 3. **Assume \( x \) is the concentration of \( \text{H}^+ \) and \( \text{CN}^- \) at equilibrium:** \[ K_a = \frac{x^2}{0.432 - x} \approx \frac{x^2}{0.432} \] 4. **Solve for \( x \):** \[ x^2 = 4.9 \times 10^{-10} \times 0.432 \] Calculate \( x \), which represents \( [\text{H}^+] \). 5. **Calculate the pH:** \[ \text{pH} = -\log[\text{H}^+] \] 6. **Find pOH using the relation:** \[ \text{pOH} = 14 - \text{pH} \] This step-by-step guide helps in determining the equilibrium pOH of a weak acid solution using its dissociation constant and initial concentration.
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