To find the equilibrium concentrations of [tex]\( \text{H}_3\text{O}^+ \)[/tex], [tex]\( \text{CN}^- \)[/tex], and HCN in a 0.011 M solution of HCN with [tex]\( K_a = 4.0 \times 10^{-10} \)[/tex], we can follow these steps:
1. Establish Initial Conditions and Define Variables:
- The initial concentration of HCN is [tex]\(0.011 \, \text{M}\)[/tex].
- The dissociation constant [tex]\( K_a \)[/tex] is [tex]\(4.0 \times 10^{-10}\)[/tex].
2. Set Up the Equilibrium Expression:
- HCN dissociates as: [tex]\[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \][/tex]
- At equilibrium, let [tex]\( [\text{H}_3\text{O}^+] = x \)[/tex] and [tex]\( [\text{CN}^-] = x \)[/tex]. The concentration of HCN at equilibrium will be [tex]\( [\text{HCN}]_{\text{initial}} - x \)[/tex].
3. Write the Expression for the Dissociation Constant [tex]\(K_a\)[/tex]:
[tex]\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} = \frac{x \cdot x}{0.011 - x} \][/tex]
4. Simplify the Equation:
- Since HCN is a weak acid, we assume [tex]\( x \)[/tex] is significantly smaller than 0.011, so [tex]\( 0.011 - x \approx 0.011 \)[/tex].
[tex]\[ K_a \approx \frac{x^2}{0.011} \][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = K_a \cdot 0.011 \][/tex]
[tex]\[ x = \sqrt{K_a \cdot 0.011} \][/tex]
- Plugging in the values: [tex]\( x = \sqrt{4.0 \times 10^{-10} \times 0.011} \)[/tex].
6. Calculate [tex]\( x \)[/tex]:
- [tex]\( x \)[/tex] is found to be approximately [tex]\( 2.097617696340303 \times 10^{-6} \)[/tex].
7. Determine the Equilibrium Concentrations:
- [tex]\( [\text{H}_3\text{O}^+] = x = 2.097617696340303 \times 10^{-6} \, \text{M} \)[/tex].
- [tex]\( [\text{CN}^-] = x = 2.097617696340303 \times 10^{-6} \, \text{M} \)[/tex].
- [tex]\( [\text{HCN}] = 0.011 - x = 0.01099790238230366 \, \text{M} \)[/tex].
8. Calculate the pH:
- The pH is defined as: [tex]\( \text{pH} = -\log_{10} [\text{H}_3\text{O}^+] \)[/tex].
- Substituting the value: [tex]\( \text{pH} = -\log_{10} (2.097617696340303 \times 10^{-6}) \)[/tex].
- This gives a pH of approximately [tex]\( 5.678273661756906 \)[/tex].
Therefore, the equilibrium concentrations and pH of the solution are:
[tex]\[ [\text{H}_3\text{O}^+] = 2.097617696340303 \times 10^{-6} \, \text{M} \][/tex]
[tex]\[ [\text{CN}^-] = 2.097617696340303 \times 10^{-6} \, \text{M} \][/tex]
[tex]\[ [\text{HCN}] = 0.01099790238230366 \, \text{M} \][/tex]
[tex]\[ \text{pH} = 5.678273661756906 \][/tex]
