What amount of HCl will be required to prepare one litre of a buffer solution of pH 10.4 using 0.01 mole of NaCN ? Given Kion HCN = 4.1 × 10-10.

The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have[NaCN] = (0.01- x)[HCN] = xSubstitution these values along with pH and Ka in the expression.pH = - log Ka + log  SaltAcidWe get 10.4 =-log  4 × 10-10 + log  0.01 -xxor 10.4 = 9.4 + log  0.01 -xxor  log  0.01 -xx=1or  0.01 -xx=10 ⇒  11x = 10-2or   x = 9.9 × 10-4 MSo, number of moles  = 9.9 × 10-4