Evaluate the pH during the titration:

i)  The pH of a solution of NaCN can be calculated as

CN ^- + H₂O ↔    HCN + OH^-

K_b = [OH^-][HCN]/[CN^-] = K_w/K_a =1.00×10^-14/6.2×10^-10 =1.61×10^-5

Sinceanequ ivalentamountof [OH^-] and [HCN] areformed

[OH^- ] = [HCN]

[CN^-] = c _NaCN - [OH^-] » c NaCN = 0.05M

Substitution into the above dissociation-constant expression provides, after rearrangement,

pH = 14.00 - (- log 8.97×10 ^-4 ) 10.95

ii)        10.00 cm³ of Reagent

Addition of acid generates a buffer along with a composition given through

c_NaCN =50.00×0.05-10.00×0.100/60.00 = 1.50/60.00 M

c_HCN = 10.00×0.10/60.00 = 1.000/ 60.00 M

Those values are then substituted within the expression for the acid dissociation constant of HCN to give [H⁺] directly:

[H⁺]=K_a×c_acid/c_salt

[H⁺] = 6.2×10^-10×(1.000/60.00)/1.50/60.00 = 4.13×10^-10

pH = -log4(4.13×10^-10)=9.38

iii)       25.00 cm³ of Reagent

This volume corresponds to the equivalence point, where the principal solution species is the weak acid HCN. Therefore,

c_HCN =25.00×0.10/75.00 =0.033M

Applying following Equation gives

pH = -log(4.45×10^-6)=5.34

iv)       26.00 cm³ of Reagent

The excess of strong acid now present represses the dissociation of the HCN to the point where its contribution to the pH is negligible. Thus,

[H⁺] = c_HCI = (26.00×0.10-50.00×0.05)/76.00 =1.32×10^-3M

pH= -log(1.32×10^-3)=2.88