Question

(Taking place in a 25 degree Celcius system)

If 40.0 mL of a 0.250 M solution of HCOOH is titrated with 0.500 M NaOH, what is the pH

a. Before the titration begins?

b. At the midpoint of the titration?

c. When 15.0 mL of NaOH is added?

d. At the equivalence point of the titration?

e. When 30.0 mL of NaOH has been added?

I already know the answers, I just really wish to understand each step and the entire process. Thank you!

Answer 1

pKa = -logKa = = -log(1.8×10^–4.) = 3.74

millimoles of HCOOH = 40 x0.250= 10

a) before titration

pH = 1/2 (pKa- log C)

   = 1/2 (3.74 -log (0.250) ) = 2.17

pH= 2.17

(b) At the midpoint of the titration

at mid point of titration , means at the half - equivalence point .

at half - equivalence point

pH = pKa

pH = 3.74

(c) after addition of 15.0 mL of NaOH

millimoles of NaOH = 15 x 0.5 = 7.5

HCOOH + NaOH ------------------------------>HCOONa + H2O

10            7.5                                         0               0 -----------------------initial

2.5         0                                          7.5 7.5 -------------------equilibirum

pH = pKa + log[salt/acid]

    = 3.74 + log (7.5 / 2.5)

    = 4.22

pH = 4.22

(d) At the equivalence point

at equivalence point :

millimoles of acid = millimoles of base

40 x 0.250 = 0.5 x V

V = 20 mL

salt only remains :

salt concentration = 10 / (40 + 20) = 0.167 M

pH = 7 + 1/2 (pKa + log C)

      = 7 + 1/2 (3.74 + log 0.167)

     = 8.48

pH = 8.48