Question

Calculate pH at equivalence point when 100 mL of a 0.1 M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 x 10^-5, is titrated with a 0.10 M NaOH solution?

Please show your ICE chart and why you chose to use the equations you did.

Answer 1

find the volume of NaOH used to reach equivalence point

M(HC2H3O2)*V(HC2H3O2) =M(NaOH)*V(NaOH)

0.1 M *100.0 mL = 0.1M *V(NaOH)

V(NaOH) = 100 mL

we have:

M(HC2H3O2) = 0.1 M

V(HC2H3O2) = 100 mL

M(NaOH) = 0.1 M

V(NaOH) = 100 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.1 M * 100 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HC2H3O2) = 10 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react to form C2H3O2- and H2O

C2H3O2- here is strong base

C2H3O2- formed = 10 mmol

Volume of Solution = 100 + 100 = 200 mL

Kb of C2H3O2- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofC2H3O2-,c = 10 mmol/200 mL = 0.05M

C2H3O2- dissociates as

C2H3O2- + H2O -----> HC2H3O2 + OH-

0.05 0 0

0.05-x x x

Kb = [HC2H3O2][OH-]/[C2H3O2-]

Kb = x*x/(c-x)

Assuming 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*10^-2) = 5.27*10^-6

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

so, x = 5.27*10^-6 M

[OH-] = x = 5.27*10^-6 M

we have below equation to be used

pOH = -log [OH-]

= -log (5.27*10^-6)

= 5.28

we have below equation to be used

PH = 14 - pOH

= 14 - 5.28

= 8.72

Answer: 8.72