Calculate the pH during the titration:

Calculate the pH during the titration of 50.00 cm³ M NaOH with 0.10 M HCl after the addition of the following volume of acid:  (i) 24.50 cm³, (ii) 25.00 cm³ and (iii) 25.50 cm³.

i)          At 24.50 cm³  added, [H⁺] is very small and cannot be computed from stoichiometric considerations but can be obtained from [OH^- ]

[OH^- ] = c_base = (Original no. mmol NaoH - No. mmol HCl added)/Total volume of solution

=50.00×0.05-24.50×0.10/50.00+24.50 = 6.71× 10^-4 M

[H⁺] = K_w /(6.71×10^-4)=1.00×10^-14 /(6.71×10-⁴)

= 1.49×10^-11 M

pH = - log(1.49×10^-11 M) =10.83

ii)        That is the equivalence point where [H⁺]= [OH^-]

pH = - log (1.00 ×10 ^-7 ) = 7.00

iii)       At 25.50 cm3 added,

[H ⁺] = cHCl = ((25.50 × 0.10 - 50.00 × 0.05)/75.50)

= 6.62 × 10^-4 M

pH = - log (6.62 ×10 ^-4 ) = 3.18