Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound A with a 0.2000 M solution of compound B in the following table. For each titration, calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45.00, 49.00, 50.00, 51.00, and 60.00 mL of com- pound B.
A
(a) H2SO3
(b) ethylenediamine
(c) H2SO4
B
NaOH
HCl
NaOH
titration between H2SO4 (from A) and NaOH from (B)
millimoles of H2SO4 = 2 x 50 x 0.1 = 10 (here 2 indicates 2 H+)
1) 0.00 ml NaOH added
H2SO4 molarity = 0.1 x 2 = 0.2
pH = -log [H+] = -log 0.2 = 0.699
2) 20 ml NaOH added
millimoles of NaOH = 20 x 0.2 = 4
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 4) / (50 +20)
= 0.0857 M
pH = 1.067
3) 24 ml NaOH added
millimoles of NaOH = 24 x 0.2 = 4.8
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 4.8) / (50 +24)
= 0.0703 M
pH = 1.153
4) 25 ml NaOH added
millimoles of NaOH = 25 x 0.2 = 5
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 5) / (50 +25)
= 0.0667M
pH = 1.176
5) 26 ml NaOH added
millimoles of NaOH = 26 x 0.2 = 5.2
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 5.2) / (50 +26)
= 0.063 M
pH = 1.20
6) 37.5 ml NaOH added
millimoles of NaOH = 37.5 x 0.2 = 7.5
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 7.5) / (50 +37.5)
= 0.02857 M
pH = 1.544
7) 20 ml NaOH added
millimoles of NaOH = 45 x 0.2 = 9
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 9) / (50 +45)
= 0.010 M
pH = 1.977
8) 49 ml NaOH added
millimoles of NaOH = 49 x 0.2 = 9.8
millimoles of H2SO4 > millimoles of NaOH
so acid dominent
[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume
= (10 - 9.8) / (50 +49)
= 2.02 x 10^-3 M
pH = 2.69
9) 50 ml NaOH added
millimoles of NaOH = 50 x 0.2 = 10
millimoles of acid = millimoles of base
so it is equivalece point . strong acid + strong base here pH= 7
10) 60 ml NaOH added
millimoles of NaOH = 60 x 0.2 = 12
millimoles of H2SO4 < millimoles of NaOH
so acid dominent
[OH] = (millimoles of NaOH - millimoles of H2SO4) / total volume
= (12 - 10) / (50 +60)
= 0.0182 M
pOH = -log [OH-]
pOH = 1.74
pH + pOH = 14
pH = 12.26
note : you see the length of the problem . for one combination we get 10 pH's for all combination just imagine this problem length. remaining combinations try yourself.
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