Question

A 50.0ml solution contains 0.120M HCN (Ka = 6.2*10^-10) and 0.240M NaCN.

a.) Calculate the pH of the solution

b.) Calculate the final pH of the solution after adding 20.0 mL of 0.0800M HCl

Answer 1

a. The solution contains a weak acid which is HCN and salt of HCN with strong base NaOH which is NaCN. So it's an acidic buffer.

Now using Henderson Heselbach equation we get

PH=PKa+log[salt]/[acid]

=-log(6.2*10^-10)+log[0.240]/[0.120]=9.207+0.3010=9.808

b) As it's a buffer solution so it will resist the PH change after addition of 20.0 ml of 0.0800 M HCl. The H+ of HCl will be taken care of by CN- produced from NaCN and thus final PH after adding 20.0 ml 0.0800 M HCl wil same as before ie 9.808