Question

a 50 ml sample of 0.095 m acetic acid (ka=1.8 x 10^-5) is being titrated with 0.106 M naoh. what is the pH at the equivalence point of the titration?

Answer 1

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.095 M *50.0 mL = 0.106M *V(NaOH)

V(NaOH) = 44.8113 mL

we have:

Molarity of CH3COOH = 0.095 M

Volume of CH3COOH = 50 mL

Molarity of NaOH = 0.106 M

Volume of NaOH = 44.8113 mL

mol of CH3COOH = Molarity of CH3COOH * Volume of CH3COOH

mol of CH3COOH = 0.095 M * 50 mL = 4.75 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.106 M * 44.8113 mL = 4.75 mmol

We have:

mol of CH3COOH = 4.75 mmol

mol of NaOH = 4.75 mmol

4.75 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 4.75 mmol

Volume of Solution = 50 + 44.8113 = 94.8113 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 4.75 mmol/94.8113 mL = 0.0501M

CH3COO- dissociates as

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

0.0501 0 0

0.0501-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*5.01*10^-2) = 5.276*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.276*10^-6 M

[OH-] = x = 5.276*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (5.276*10^-6)

= 5.28

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.28

= 8.72

Answer: 8.72