In: Chemistry
You wish to titrate a 0.193 M solution of HCN (pKa = 9.24) with a 0.183 M solution of NaOH. What is the pH of the HCN solution before the titration?
HCN before the titration:
HCN <-> H+ +CN-
Ka= [H+][CN-]/[HCN]
in equilibrium:
[H+] = x
[CN-] = x
[HCN] = M-x = 0.193 - x
so
Ka = 10^-pKA = 10^-9.24 = 5.75*10^-10
so, substitute in Ka expression:
5.75*10^-10= x*x/(0.193-x)
solve for x:
x^2 + Ka*x - M*Ka = 0
x = 1.053*10^-5
[H+] = 1.053*10^-5
pH = -log(1.053*10^-5) = 4.977
approx, before any base is added