Question

5. Consider the titration of 100.0 mL of 0.100 M H2NNH>/hydrazine (Ky = 3.0 x 10-6) by 0.200 M HNO3. Assume that hydrazine is monoprotic. Determine the following: a. pH before any HNO3 is added b. volume of base to reach the equivalence point c. pH when 20.0 mL of HNO, has been added d. pH when 25.0 mL of HNO, has been added e. pH when 40.0 mL of HNO3 has been added f. pH when 50.0 mL of HNO3 has been added g. pH when 100.0 mL of HNO3 has been added

please show all work and write neatly. when solving the 50ml and 100ml please explain how to find the ph.

Answer 1

(a) Before HNO₃ added

H2NNH2 is a weak base :
H2NNH2 + H2O <----> H2NNH3+ + OH-
initial concentration
0.1
at equilibrium
0.1-x . . . . . . . .. . . .. . . . . . . x . . . .. .. .x
3.0 x 10^-6 = x² / 0.1-x
x = [OH-] = 0.000548 M
pOH = - log 0.000548 = 3.26
pH = 14 - pOH

pH = 10.74

(b) At equivalence point

pH = pK_a = 14 - pkb = 14 - 5.52 = 8.48

(c) 20.0 mL

Moles HNO3 = 0.200 x 20 /1000 = 0.004
Moles H2NNH2 = 0.1 x 100 /1000 = 0.01
H+ + H2NNH2 >> H2NNH3+
0.01 mole H2NNH2 + 0.004 mole H+ >> 0.006 mole H2NNH2 in excess + 0.004 mole H2NNH3+
Total volume = 120 mL = 0.120 L
[H2NNH2 ] = 0.006 / 0.120 = 0.05 M
[H2NNH3+] = 0.004 / 0.120 = 0.03 M
3.0 x 10^-6 ( 0.05 / 0.03 ) = [OH-] = 5 x 10^-6 M
pOH = 5.3
pH = 14-pH = 14-5.3

pH = 8.70

(d) 25.0 mL

Moles H2NNH2 = 0.1 x 100 /1000 = 0.01
Moles HNO3 = 0.2 x 25 /1000 = 0.005
0.01 - 0.005 = 0.005 mole H2NNH2 in excess
Mole H2NNH3+ = 0.005
Total volume = 125 mL = 0.125 L
[H2NNH2] = 0.005 / 0.125 = 0.04 M
[ H2NNH3+ ] =0.005 / 0.125 = 0.04 M
[OH-] = 3.0 x 10^-6 ( 0.04 / 0.04 ) = 3 x 10^-6 M
pOH = 5.52
pH = 8.47

(e) 40 mL
Moles H2NNH2 = 0.1*100/1000=0.01
Moles HNO3 = 0.2 x 40 /1000 = 0.008
0.01 - 0.008 = 0.002 mole H2NNH2 excess
mole H2NNH3+ = 0.008
Total volume = 0.140 L
[ H2NNH2] = 0.01428 M
[ H2NNH3+] = 0.05714 M
[OH-] = 3.0 x 10^-6 ( 0.01428 / 0.05714 ) = 0.74 x 10^-6 M
pOH = 6.13
pH = 7.87

(1) 50.0 mL At this point, all H,NNH, is converted into H,NNH, a weak acid. The pH is calculated by considering the acid dissociation reaction of H NNH HANNH; = H,NNH, +H* F-* * * F = (0.100 M)_100.0_) 100.0 +50.0 = 0.0667 M [H][H,NNH,] (H2NNH:] 1.0 x 10-24 3.0 x 10-6 - = 3.3 10- 0.0667-X x = 1.5 x 10- Therefore, pH = -log[ht] =-log(1.5 10-) = 4.82

(g) 100.0 mL After the equivalence point, the excess of strong acid determines the pH. [v+] = (0.200 M)(1000+1000) = 0.0500 M Therefore, pH = -log[H] =-log(0.0500) = 1.30