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The pKa of HCN is 9.3. The pH of a solution prepared by mixing 2.5 moles of HCN and 0.25 moles of KCN in water and making of the total volume to 500 ml is:
A. 9.3
B. 7.3
C. 10.3
D. 8.3

Answer
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Hint: There is a formula to calculate the pH of the solution by using pKa and it is as follows.
pH=pKa+log[salt][acid]
pH = pH of the solution
pKa = acid ionization constant
[salt] = concentration of the salt
[acid] = concentration of the acid

Complete step by step solution:
- In the question it is given that 2.5 moles of HCN and 0.25 moles of KCN in water and made the solution to 500 ml by using water.
- We have to calculate the pH of the resulting solution.
- It is given that Acid ionization constant pKa = 9.3
- Concentration of the salt (KCN) = 0.25 moles
- Concentration of the acid (HCN) = 2.5moles
- Substitute all the known values in the below formula to get the pH of the resulting solution.
pH=pKa+log[salt][acid]
pH = pH of the solution
pKa = acid ionization constant = 9.3
[salt] = concentration of the salt = 0.25 moles
[acid] = concentration of the acid = 2.5 moles
pH=pKa+log[salt][acid]pH=9.3+log0.252.5pH=8.3
- Therefore the pH of the resulting mixture is 8.3.

So, the correct option is D.

Note: HCN (hydrocyanic acid) is a weak acid and KCN (potassium cyanide) is salt. By mixing these two we will get an acidic buffer.
KCN is going to be formed by the reaction of HCN with KOH and it is as follows.
HCN+KOHKCN+H2O