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.)
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}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1cf2ec1-831d-43ac-84b4-19fe406513bb%2F1c0f0e05-bf6d-44cc-b1d8-b7c95544521b%2Fltqy80r_processed.jpeg&w=3840&q=75)
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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