Consider the following equilibrium reaction for hydrocyanic acid, HCN. HCNHH* +CN- At 25°C the concentrations are found to be [HCN] = 0.0160 M and [H+] = [CN-] = 2.0 x 10-6 M. What is the value of the equilibrium constant at 25°C? (Show your work.)

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### Equilibrium Constant Calculation for Hydrocyanic Acid

Consider the following equilibrium reaction for hydrocyanic acid, HCN:

\[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \]

At 25°C the concentrations are found to be:
- \([\text{HCN}] = 0.0160 \; \text{M}\)
- \([\text{H}^+] = [\text{CN}^-] = 2.0 \times 10^{-6} \; \text{M}\)

**Question:**
What is the value of the equilibrium constant at 25°C? (Show your work.)

#### Solution:

For the equilibrium reaction:

\[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \]

The equilibrium constant expression (\(K_c\)) is given by:

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

Given the concentrations:
\[ [\text{HCN}] = 0.0160 \; \text{M} \]
\[ [\text{H}^+] = 2.0 \times 10^{-6} \; \text{M} \]
\[ [\text{CN}^-] = 2.0 \times 10^{-6} \; \text{M} \]

Substitute these values into the equilibrium constant expression:

\[ K_c = \frac{(2.0 \times 10^{-6} \; \text{M})(2.0 \times 10^{-6} \; \text{M})}{0.0160 \; \text{M}} \]

Calculate the product of the concentrations of \(\text{H}^+\) and \(\text{CN}^-\):

\[ (2.0 \times 10^{-6} \; \text{M}) \times (2.0 \times 10^{-6} \; \text{M}) = 4.0 \times 10^{-12} \; \text{M}^2 \]

Now divide by the concentration of \(\text{HCN}\):

\[ K_c = \frac{4.0 \times 10^{-12}
Transcribed Image Text:### Equilibrium Constant Calculation for Hydrocyanic Acid Consider the following equilibrium reaction for hydrocyanic acid, HCN: \[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \] At 25°C the concentrations are found to be: - \([\text{HCN}] = 0.0160 \; \text{M}\) - \([\text{H}^+] = [\text{CN}^-] = 2.0 \times 10^{-6} \; \text{M}\) **Question:** What is the value of the equilibrium constant at 25°C? (Show your work.) #### Solution: For the equilibrium reaction: \[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \] The equilibrium constant expression (\(K_c\)) is given by: \[ K_c = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \] Given the concentrations: \[ [\text{HCN}] = 0.0160 \; \text{M} \] \[ [\text{H}^+] = 2.0 \times 10^{-6} \; \text{M} \] \[ [\text{CN}^-] = 2.0 \times 10^{-6} \; \text{M} \] Substitute these values into the equilibrium constant expression: \[ K_c = \frac{(2.0 \times 10^{-6} \; \text{M})(2.0 \times 10^{-6} \; \text{M})}{0.0160 \; \text{M}} \] Calculate the product of the concentrations of \(\text{H}^+\) and \(\text{CN}^-\): \[ (2.0 \times 10^{-6} \; \text{M}) \times (2.0 \times 10^{-6} \; \text{M}) = 4.0 \times 10^{-12} \; \text{M}^2 \] Now divide by the concentration of \(\text{HCN}\): \[ K_c = \frac{4.0 \times 10^{-12}
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