Question

What is the pH at the equivalence piont in the titration of 25 mL of .10 M HOAc with .1 M NaOH? Ka= 1.8x10^-5?

Please show work?

Answer 1

CH3COOH millimoles = 0.1 x25 = 2.5

NaOH millimoles = 0.1 x 2.5 = 2.5

CH3COOH + NaOH -----------------------> CH3COONa    + H2O

2.5                   2.5                                      0                      0       --------------------initial

0                      0                                          2.5                 2.5 -------------------equilibrium

in this mixture contain only salt CH3COONa

salt concentration = millimoles/total volume

                             = 2.5/50

                             = 0.05

now we have to use salt hydrolysis concept

CH3COO- + H2O ---------------------------> CH3COOH + OH-

0.05                                                              0                    0       ------------------initial

0.05-x                                                           x                   x ----------------------equlibrium

Kb = [CH3COOH][OH-]/[CH3COO-]

   Kw/Ka    = x^2/(0.05-x)

1 x10^-14/1.8x10^-5 = x^2/(0.05-x)

5.5 x 10^-10 = = x^2/(0.05-x)

x^2 + 5.5 x 10^-10 .x - 0.275 x 10^-10 = 0

x = 5.26 x 10^-6

[OH-] = x = 5.26 x 10^-6 M

pOH = -log[OH-] = -log( 5.26 x 10^-6) = 5.27

pH + pOH = 14

pH = 14-POH = 14-5.72

pH = 8.72