acid & base

Assume that V dm3 of 0.105 M NaOH(aq) and V dm3 of 0.095 M HCl(aq) are mixed.

OH- + H+ → H2O
Mole ratio OH- : H+ = 1 : 1

No. of moles of OH-(aq) added = 0.105V dm3.
No. of moles of H+(aq) added = 0.095V dm3.
OH-(aq) is in excess, and H+(aq) is the limiting reactant.

No. of moles of unreacted OH-(aq) = 0.105V - 0.095V = 0.01V mol
Volume of the final solution = V + V = 2V dm3
[OH-] of the final solution = (0.01V)/(2V) = 0.005 mol dm-3
[H+] of the final solution = Kw/[OH-] = (1 x 10-14)/0.005 = 2 x 10-12
pH of the final solution = -log[H+] = -log(2 x 10-12) = -11.7
=