An aqueous buffer solution contains only HCN (PK, = 9.31) and KCN and has a pH of 8.50. What can be concluded about the relative concentrations of HCN and KCN in the buffer?
(A) [HCN]> [KCN]
(B) [HCN]< [kcn]=””>
(C) [HCM] = [KCN]
(D) nothing can be concluded about the relative concentrations
The Correct Answer and Explanation is:
Correct Answer: (A) [HCN] > [KCN]
Explanation:
The given buffer solution contains hydrogen cyanide (HCN), a weak acid, and potassium cyanide (KCN), its conjugate base. KCN dissociates completely in water to form CN⁻, the conjugate base of HCN.
To analyze the relative concentrations of HCN and CN⁻ (from KCN), we use the Henderson-Hasselbalch equation: pH=pKa+log([Base][Acid])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
Here:
- pH = 8.50 (given)
- pKₐ = 9.31 (for HCN)
- Base = CN⁻ (from KCN)
- Acid = HCN
Substituting into the equation: 8.50=9.31+log([CN−][HCN])8.50 = 9.31 + \log\left(\frac{[\text{CN}^-]}{[\text{HCN}]}\right)
Solving for the log term: log([CN−][HCN])=8.50−9.31=−0.81\log\left(\frac{[\text{CN}^-]}{[\text{HCN}]}\right) = 8.50 – 9.31 = -0.81
Now convert the log value: [CN−][HCN]=10−0.81≈0.155\frac{[\text{CN}^-]}{[\text{HCN}]} = 10^{-0.81} \approx 0.155
This means: [CN−]<[HCN][\text{CN}^-] < [\text{HCN}]
Since CN⁻ comes from KCN, we can conclude that [KCN] < [HCN].
Conclusion:
In the buffer, the pH (8.50) is lower than the pKₐ (9.31), which means the solution is more acidic than the point at which equal concentrations of acid and base would occur. Therefore, the concentration of the acid (HCN) must be greater than the concentration of the base (CN⁻ from KCN).
Thus, the correct answer is (A) [HCN] > [KCN].
