Question

A 25.00 mL sample containing a certain monoprotic weak acid, HA (K_a 4.5 x 10^-5) requires 16.25 mL of 0.102 M NaOH to reach the endpoint of titration.

a) Determine the concentration of the acid in the sample.

b) Determine the pH of the original solution before any titrant was added.

c) Determine the concentration of the conjugate base A^- at the endpoint of titration.

d) Determine the pH of the solution at the endpoint of the titration.

Answer 1

a)

HA + NaOH = NaA + H2O

then, verify the stoichiometric ratio of acid : base

it is 1:1

then

mmol of acid = mmol of base

mmol of base = MV = (0.102)(16.25) = 1.6575 mmol of base

therefore,

mmol of acid = 1.6575 mmol

[HA]= mmol of HA / V of acid = 1.6575 / (25) = 0.0663 M of HA

b)

original solution of acid:

HA = H+ + A-

Ka = [H+][A-]/[HA]

initially

[H+] = 0

[A-] = 0

[HA] = 0.0663

in equilibrium:

[H+] = 0 + x

[A-] = 0 + x

[HA] = 0.0663 - x

substitute in Ka

Ka = [H+][A-]/[HA]

4.5*10^-5 = (x*x)/(0.0663-x)

solve for x

x = 0.0017

[H+] = x = 0.0017

now,

pH = -log(H+) = -log(0.0017) = 2.77

c)

in the endpoint:

Vtotal = Vacid + Vbase = 25+16.25 = 41.25 mL

now,

we know that all HA will transform to A- so

1.6575 mmol of HA = 1.6575 mmol of A-

then

[A-] = mmol of A- / Vtotal = 1.6575 / 41.25 = 0.040181 M

d)

find pH in endpoint...

A- is present so this will form hydrolysis

A-. + H2O = HA + OH-

Kb = [HA][OH-]/[A-]

Kb can be calcualted via

Kw = Ka*Kb

Kb = Kw/Ka = (10^-14)/(4.5*10^-5) = 2.2222*10^-10

then, similar as in b:

Kb = [HA][OH-]/[A-]

2.222*10^-10 = x*x/(0.040181-x)

x = OH- = 2.986*10^-6

pOH = -log(2.986*10^-6) = 5.5249

pH = 14-pOH = 14-5.5249

pH = 8.4751, basic as expected