posted by Amandeep on .
What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10?
Rate my answer please
For your understanding:
HCN ⇌ H+ + CN-,
Ka = 6.2x10^-10 = [H+]*[CN-]/[HCN]
Notice the H+ and CN- ions are generated in pairs during HCN dissociation, hence [H+] = [CN-]. Hence:
[H+]^2 = 6.2x10^-10*[HCN] = 7.688x10^-10
pH = -log([H+]) = -0.5*log(7.688x10^-10) = 4.56
1st. It would help for understanding if you showed next to last step as
[H^+]^2= 6.2 x 10^-10*1.24 = 7.688 x 10^-10.
2nd. I'm not sure how clear the last step is. I prefer to take the square root of 6.2 x 10^-10 = ??, THEN do the pH = -log(H^+). The way you have done it the reader must recognize that you took the log of the squared term so you must divide by 2.