Question

Titration of a strong base with a strong acid

A 20.00mL sample of 0.400 M barium hydroxide is titrated with 0.200 M Hydrobromic acid. Determine the pH Of the original base. Next, what is the pH 10.00mL short of the equivalence?

Answer 1

Ba(OH)2 + 2 HBr ---------------> BaBr2 + 2H2O

20x0.4 ----- ------- -------- initial mmoles

a)

pH of base

As Ba(OH)2 is diacidic base ,

[OH-] = 2 x molarity = 2x0.4=0.8 0M

thus pOH of base = -loog [OH-] = -log0.80 =0.09691

pH of solution = 14-pOH = 13.90

b) 10.00mL short of equivalence.

At equivalence

M1V1/n1 = M2V2/n2

20mL x 0.4 /1 = 0.2 x V2/ 2

Thus volume of HBr at equivalence = 80 mL

!0mL short means

pH after addition of 70mL of HBr

Ba(OH)2 + 2 HBr ---------------> BaBr2 + 2H2O

20x0.4x2=1.6 70x0.2=1.4 ------- -------- initial mmoles

0.2 0 1.4 --- after rxn

thus the solution still has excess base and thus

[OH-] = mmoles/ volume = 0.2/(20+70) =0.0022 M

and pOH = -log [OH-] = -log 0.0022 = 2.653

and pH = 14-pOH = 14-2.653 =11.347