Question

In: Chemistry

What is the pH at the equivalence piont in the titration of 25 mL of .10...

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?

Solutions

Expert Solution

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


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