Question

Find the pH during the titration of 20.00 mL of 0.1630 M benzoic acid, C₆H₅COOH (K_a = 6.3 ? 10^-5), with 0.1630 M NaOH solution after the following additions of titrant.

(a)    0 ml

(b)    10.00 mL

(c)    15.00 mL

(d)    19.00 mL

(e)    19.95 mL

(f)    20.00 mL

(g)    20.05 mL

(h)    25.00 mL

Answer 1

Since you are titrating 20.00 mL of 0.1630 M benzoic acid with an equal strength of NaOH, and they react in a 1:1 mole ratio, then the endpoint of the titration will be at 20.00 mL of the NaOH (where moles NaOH added = moles benzoic acid). At points in between 0 and 20 mL of NaOH (i.e., during the titration), you have a buffer system because both benzoic acid and benzoate ion (the conjugate base) are present.

C6H5COOH + OH- ==> H2O + C6H5COO-

Buffer system pH's are easily calculated using the Henderson-Hasselbalch equation:

pH = pKa + log (moles A- / moles HA)

where HA is the acid species in the buffer and A- is its conjugate base. In our problem, HA = C6H5COOH and A- = C6H5COO-.

So it just becomes a matter of calculating moles of HA and A- for each volume between 0 and 20 mL.

(b) Initial moles C6H5COOH = M C6H5COOH x L C6H5COOH
= (0.1630)(0.0200) = 0.00326 moles benzoic acid

moles NaOH added = M NaOH x L NaOH = (0.0.1630)(0.0100)
= 0.00163 moles NaOH.

moles C6H5COOH remaining = 0.00326 - 0.00163 = 0.00163

moles C6H5COO- formed = moles NaOH added = 0.00163

pH = pKa + log (moles A- / moles HA)
= -log (6.3 x 10^-5) + log (0.001630 / 0.00163)
= 4.20 + log 1
= 4.20 + 0
= 4.20 ... your answer is right. At the halfway point of any titration,
pH = pKa.

ot to ignore the -x term, divide the original salt concentration (0.0725) by Kb (1.6 x 10^-10). If the result is > 100, you can ignore the x term. In our case, the ratio is 450 million!

1.6 x 10^-10 = x^2 / 0.0725
1.2 x 10^-11 = x^2
3.4 x 10^-6 = x = [OH-]

pOH = -log [OH-] = -log (3.4 x 10^-6) = 5.47

pH = 14.00 - pOH = 14.00 - 5.47 = 8.53

(g) After the endpoint, the pH is governed by the amount of strong base (NaOH) added. The amount produced by the benzoate ion is negligible.

moles NaOH added after endopoint = 20.05 mL - 20.00 mL = 0.05 mL.
moles NaOH = M NaOH x L NaOH = (0.1450)(0.000050) = 7.3 x 10^-6
The volume of the solution is 40.05 mL.

[OH-] = moles OH- / L = 7.3 x 10^-6 / 0.04005 = 1.8 x 10^-4 M

pOH = -log [OH-] = -log (1.8 x 10^-4) = 3.74

pH = 14.00 - pOH = 14.00 - 3.74 = 10.26

(h) like (g)

mL NaOH added after endpoint = 25.00 - 20.00 = 5.00 mL
moles NaOH = (0.1450)(0.0050) = 7.3 x 10^-4 moles
The solution volume is now 45.00 mL.

[OH-] = moles OH- / L = 7.3 x 10^-4 / 0.0450 = 0.0162 M

pOH = -log [OH-] = -log (0.0162) = 1.79

pH = 14.00 - pOH = 14.00 - 1.79 = 12.21

the others also are done with same process.