Question

A. 100ml of 0.5M acetic avid is titration with 20ml of 0.5M KOH. What is the pH of the acid before titration? (ka acetic acid is 1.8 x 10^-5)

B. What is the pH of the solution after titration?

Answer 1

Solution :-

A) Acetic acid is the weak acid therefore it do not dissociate 100 %

lets calculate the concnetration of the H3O+ using the Ka

CH3COOH + H2O   ----- > H3O^+   + CH3COO^-

0.5                                     0                0

-x                                       +x               +x

0.5-x                                    x               x

ka = [H3O+][CH3COO-]/[CH3COOH]

1.8*10^-5 = [x][x]/[0.5-x]

since the Ka is very small therefore we can neglect the x from the denominator

then we get

1.8*10^-5 = [x][x]/[0.5]

1.8*10^-5 * 0.5 = x^2

9.0*10^-6 = x^2

taking square root of the both sides we get

3.0*10^-3 = x

now lets calculate the pH

pH= -log [H3O+]

pH= - log [3.0*10^-3]

pH= 2.52

Therefore pH before addition of the KOH is 2.52

B) pH after adding 20.0 ml of 0.5 K KOH

lets first calculate the moles of acetic and KOH

moles = molarity * volume

Moles of acetic acid = 0.5 mol per L * 0.100 L =0.05 mol

moles of KOH = 0.5 mol per L * 0.020 L = 0.01 mol KOH

moles of the KOH are limiting therefore after the reaction 0.01 mol acetic acid is reduced and 0.01 mol acetate ion will be formed

so new moles of acetic acid = 0.05 mol - 0.01 mol = 0.04 mol

moles of acetate ion = 0.01 mol

now lets calculate the new molarities at total volume (100.0 ml + 20.0 ml = 120.0 ml = 0.120 L)

acetic acid = 0.04 mol / 0.120 L = 0.3333 M

acetateion = 0.01 mol / 0.120 L = 0.08333 M

now lets calculate the pH of the solution using the Henderson equation

pH= pka + log ([base]/[acid])

pka = -log ka

pka of acetic acid = - log 1.8*10^-5 = 4.74

lets put the values in the formula

pH= pka + log ([base]/[acid])

pH= 4.74 + log [0.0833 / 0.3333]

pH= 4.74 +(-0.602)

pH= 4.14

Therefore pH after adding the KOH = 4.14