Question

Consider the titration of 25.0 mL of 0.210 MHCO2H with 0.250 MNaOH. (Ka of HCO2H is 1.8×10^−4)

a) How many milliliters of base are required to reach the equivalence point?

b) Calculate the pH after the addition of 8.0 mL of base.

c) Calculate the pH at halfway to the equivalence point.

d) Calculate the pH after the addition of 30.0 mL of base.

Answer 1

a)

find the volume of NaOH used to reach equivalence point

M(HCO2H)*V(HCO2H) =M(NaOH)*V(NaOH)

0.21 M *25.0 mL = 0.25M *V(NaOH)

V(NaOH) = 21 mL

Answer: 21 mL

b)

1)when 8.0 mL of NaOH is added

Given:

M(HCO2H) = 0.21 M

V(HCO2H) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 8 mL

mol(HCO2H) = M(HCO2H) * V(HCO2H)

mol(HCO2H) = 0.21 M * 25 mL = 5.25 mmol

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

mol(NaOH) = 0.25 M * 8 mL = 2 mmol

We have:

mol(HCO2H) = 5.25 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

excess HCO2H remaining = 3.25 mmol

Volume of Solution = 25 + 8 = 33 mL

[HCO2H] = 3.25 mmol/33 mL = 0.0985M

[HC2-] = 2/33 = 0.0606M

They form acidic buffer

acid is HCO2H

conjugate base is HC2-

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.745+ log {6.061*10^-2/9.848*10^-2}

= 3.534

Answer: 3.53

c)

At half equivalence point,

pH = pKa

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

So,

pH = 3.745

Answer: 3.75

d)

Given:

M(HCO2H) = 0.21 M

V(HCO2H) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 30 mL

mol(HCO2H) = M(HCO2H) * V(HCO2H)

mol(HCO2H) = 0.21 M * 25 mL = 5.25 mmol

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

mol(NaOH) = 0.25 M * 30 mL = 7.5 mmol

We have:

mol(HCO2H) = 5.25 mmol

mol(NaOH) = 7.5 mmol

5.25 mmol of both will react

excess NaOH remaining = 2.25 mmol

Volume of Solution = 25 + 30 = 55 mL

[OH-] = 2.25 mmol/55 mL = 0.0409 M

use:

pOH = -log [OH-]

= -log (4.091*10^-2)

= 1.3882

use:

PH = 14 - pOH

= 14 - 1.3882

= 12.6118

Answer: 12.61