Question

Calculate the pH at the stoichiometric point when 25 mL of 0.092 M hydrazine is titrated with 0.30 M HCl.

Calculate the pH at the stoichiometric point when 25 mL of 0.080M formic acid is titrated with 0.32 M NaOH.

Calculate the pH at the stoichiometric point when 75 mL of 0.098 M hydrochloric acid is titrated with 0.33 M NaOH.

Answer 1

the Kb of hydrazine = 1.3*10^-6

which is a base

so

B + H2O <-> HB+ + OH-

for titration

B + H+ ---> BH+

in equilibrium, BH will hydorlyse:

BH+ + H2O <-> B + H3O+

in stoichiometric point:

Ka = [B][H3O+]/[BH+]

Ka= Kw/Kb = (10^-14)/(1.3*10^-6) = 7.69230*10^-9

so

7.69230*10^-9 = x*x/(M-x)

find concnetration in equivalence point:

mmol of base= MV = 25*0.092 = 2.3

mmol of acid = MV --> V = mmol/M = 2.3/(0.3) = 7.666 mL

total V = 7.666 + 25 = 32.666

so

[BH+] = mmol/total V = 2.3 /32.666= 0.0704096

7.69230*10^-9 = x*x/(0.0704096-x)

[H3O+] = x = 2.32*10^-5

pH = -log( 2.32*10^-5) = 4.634

b)

Formic acid will hydolyze:

HA <--> +H+ + OH- + A- --< H2O + A-

A- + H2O <-> HA + OH-

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

calculate total V:

mmol of acid = MV = 25*0.08 = 2

Vbase = mmol/M = 2/0.32 = 6.25 mL

total V = 6.25+25 = 31.25 mL

so

[A-] = mmol/V = 2/31.25 = 0.064 M

so.

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

Kb = Kw/Ka = (10^-14)/(1.8*10^-4) = 5.555*10^-11

5.555*10^-11 = x*x/(0.064 -x)

x = 1.8846*10^-6

pOH = -log(1.8846*10^-6) = 5.72478

pH = 14-5.72478 = 8.27522

finally:

HCl + NaOH is strong acid/base

so pH = 7 due to neutralization