Question

consider the titration of a strong acid, HCL, with a strong base NaOH. The 35 ml of .1 M HCL is in the flask. What is the pH of the solution after 40 ml of .100 M NaOH has been added to to it?

Answer 1

The reaction is,    HCl + NaOH ------->   NaCl + H₂O

So for complete neutralisation HCl and NaOH must be in equimolar amounts.

Before mixing, [HCl] = 0.1M = [NaOH] before mixing.

V_NaOH =40ml     V_HCl = 35ml       V_NaOH > V_HCl

Since concentrations of NaOH and HCl solutions are same but volume of NaOH added is greater than volume of HCl, the resulting mixture will be basic due to excess of NaOH.

Total volume is V_Final = V_NaOH + V_HCl = 40 + 35 = 75ml

We can find molarity of NaOH in final solution, M_Final using the following equation,

M_Final = {(MV)_NaOH - (MV)_HCl} / V_Final

M_Final = {(0.1*40) - (0.1*35)} / 75    =    {4 - 3.5} / 75 = 6.667*10^-3 M

Hence, concentration of NaOH in final solution is M_Final = 6.667*10^-3 M

Therefore, in final solution [OH^-] = 6.667*10^-3 M or 6.667*10^-3 mol/dm³

pOH of the solution is given by, pOH = -log[OH^-]

pOH = -log 6.667*10^-3 = log (6.667*10^-3)^-1 =log (1 / 6.667*10^-3 ) = log149.992 = 2.18

Therefore pOH of the solution is 2.18.

We know that, pH + pOH = 14

So, pH = 14 - pOH

pH = 14 - 2.18 = 11.82

So pH of the solution is 11.82