Question

What is the concentration of Ba(CN)₂ in the aqueous solution of the mixture of Ba(CN)₂ and 0.1 M HCN having pH = 5 ? Given; pKb for CN⁻ at 25°C is 8 :- don't spam answer correctly only if you know.
I will mark the perfect answer with exclamation as brainliest.

Answer 1

Given:

Molarity of HCN = 0.1 M

pkb for CN⁻ 8

Temperature T = 25°C

To find:

concentration of Ba(CN)₂ =?

Step-to-step-explanation:

• The mixture of Ba(CN)₂ in an aqueous solution dissolves resulting in the formation of BA⁺ ion &  CN⁻.
• We have to find the concentration of  Ba(CN)₂ in the aqueous solution of the mixture of Ba(CN)₂ and 0.1 M HCN.
• The pH of the given solution is 5.
• we know that

         pka + pkb = 14           ...(1)

• Let C be the required concentration.

• The pH of the mixture in the aqueous solution is given by

pH = 0.5 [ pKw + pka + log C]

pH = 0.5 [ pKw + (14 - pkb) + log C]             ...from(2)

5 = 0.5 [ 14 + (14 -5) + log C]

5 = 0.5 [ 14 + 9 + log C]

2.5 = 23 + log C

2.5 -23 = log C

- 20.5 = log C

C = 0

• This shows that the mixture dissolves in the aqueous solution.

• Hence the concentration of  Ba(CN)₂ in the aqueous solution of the mixture of Ba(CN)₂ and 0.1 M HCN is 0.