Question

What is the hydronium ion concentration in a solution prepared by mixing 150.00 mL of 0.10 MHCN with 150.00 mL of 0.030 MNaCN? Assume that the volumes of the solutions are additive and that Ka=4.9×10−10.

Answer 1

Ka=4.9×10−10.

The mixture of HCN and NaCN is a buffer thusfirst calculate the pH of this solution as follows:

pH = pKa + log [NaCN] / [HCN]

pKa = -log Ka

pKa = - log [4.9 x 10^-10]

pKa = 9.3

total volume = 150++150= 300 ml

number of moles = moalrity * volumes in l

and molaroty after new volume = number of mole s/ volume in L

[HCN] = 150 x 0.03 / 300 = 0.015 M

[HCN] = 150 x 0.1 / 300 = 0.05 M

Therefore,

pH = 9.3 + log [0.015] / [0.05]

pH = 8.78

Given the pH, we can find the hydronium concentration by taking the anti-log of 8.78

[H3O+] = 10^-8.78

= 1.66x10^-9 M is the solution