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### 1. Write the given data and the chemical equation:
The chemical equation for the dissociation of HCN is:
[tex]\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \][/tex]
The equilibrium constant [tex]\(K_c\)[/tex] for this reaction is given as:
[tex]\[ K_c = 4.9 \times 10^{-10} \][/tex]
The initial concentration of HCN is:
[tex]\[ [\text{HCN}]_0 = 0.150 \text{ M} \][/tex]
### 2. Set up the initial and equilibrium concentrations:
Initially:
- [tex]\([ \text{HCN} ]_0 = 0.150 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_0 = 0 \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_0 = 0 \text{ M}\)[/tex]
At equilibrium, if [tex]\(x\)[/tex] is the amount of HCN that dissociates:
- [tex]\([ \text{HCN} ]_{eq} = 0.150 - x \)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} = x \)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} = x \)[/tex]
### 3. Write the equilibrium expression:
The equilibrium constant expression for this reaction is:
[tex]\[ K_c = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \][/tex]
Substitute the equilibrium concentrations into this expression:
[tex]\[ 4.9 \times 10^{-10} = \frac{(x)(x)}{0.150 - x} \][/tex]
### 4. Solve for [tex]\(x\)[/tex]:
This produces a quadratic expression:
[tex]\[ 4.9 \times 10^{-10} = \frac{x^2}{0.150 - x} \][/tex]
### 5. Simplify and solve the quadratic equation:
Given the numerical solution from the software or advanced calculations, the value of [tex]\(x\)[/tex] (concentration of [tex]\( \text{H}^+ \)[/tex] at equilibrium) is:
[tex]\[ x \approx 8.57 \times 10^{-6} \text{ M} \][/tex]
### 6. Calculate the equilibrium concentrations of all species:
- [tex]\([ \text{HCN} ]_{eq} = 0.150 - x \approx 0.150 - 8.57 \times 10^{-6} \approx 0.14999143 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} = x \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} = x \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
So, the equilibrium concentrations are:
- [tex]\([ \text{HCN} ]_{eq} \approx 0.14999143 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
Thus:
[tex]\[ [ \text{HCN} ]_{eq} \approx 1.4999143 \times 10^{-1} \text{ M} \][/tex]
To match your format of the answer:
[tex]\[ [ \text{HCN} ]_{eq} = 1.49991 \times 10^{-1} \text{ M} \][/tex]
Feel free to let me know if there are other parts of the problem or specific clarifications you need!
### 1. Write the given data and the chemical equation:
The chemical equation for the dissociation of HCN is:
[tex]\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \][/tex]
The equilibrium constant [tex]\(K_c\)[/tex] for this reaction is given as:
[tex]\[ K_c = 4.9 \times 10^{-10} \][/tex]
The initial concentration of HCN is:
[tex]\[ [\text{HCN}]_0 = 0.150 \text{ M} \][/tex]
### 2. Set up the initial and equilibrium concentrations:
Initially:
- [tex]\([ \text{HCN} ]_0 = 0.150 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_0 = 0 \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_0 = 0 \text{ M}\)[/tex]
At equilibrium, if [tex]\(x\)[/tex] is the amount of HCN that dissociates:
- [tex]\([ \text{HCN} ]_{eq} = 0.150 - x \)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} = x \)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} = x \)[/tex]
### 3. Write the equilibrium expression:
The equilibrium constant expression for this reaction is:
[tex]\[ K_c = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \][/tex]
Substitute the equilibrium concentrations into this expression:
[tex]\[ 4.9 \times 10^{-10} = \frac{(x)(x)}{0.150 - x} \][/tex]
### 4. Solve for [tex]\(x\)[/tex]:
This produces a quadratic expression:
[tex]\[ 4.9 \times 10^{-10} = \frac{x^2}{0.150 - x} \][/tex]
### 5. Simplify and solve the quadratic equation:
Given the numerical solution from the software or advanced calculations, the value of [tex]\(x\)[/tex] (concentration of [tex]\( \text{H}^+ \)[/tex] at equilibrium) is:
[tex]\[ x \approx 8.57 \times 10^{-6} \text{ M} \][/tex]
### 6. Calculate the equilibrium concentrations of all species:
- [tex]\([ \text{HCN} ]_{eq} = 0.150 - x \approx 0.150 - 8.57 \times 10^{-6} \approx 0.14999143 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} = x \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} = x \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
So, the equilibrium concentrations are:
- [tex]\([ \text{HCN} ]_{eq} \approx 0.14999143 \text{ M}\)[/tex]
- [tex]\([ \text{H}^+ ]_{eq} \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
- [tex]\([ \text{CN}^- ]_{eq} \approx 8.57 \times 10^{-6} \text{ M}\)[/tex]
Thus:
[tex]\[ [ \text{HCN} ]_{eq} \approx 1.4999143 \times 10^{-1} \text{ M} \][/tex]
To match your format of the answer:
[tex]\[ [ \text{HCN} ]_{eq} = 1.49991 \times 10^{-1} \text{ M} \][/tex]
Feel free to let me know if there are other parts of the problem or specific clarifications you need!
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