A) Describe the acid-base characteristics of regions B, D and F and points A, C

Question

A) Describe the acid-base characteristics of regions B, D and F and points A, C and E. (B) Suppose that a titration vessel contains 100.00 ml of 0.100M H2B (K1 = 1.00 x 10-4 and K2 = 1.00 x 10 –7) and it is titrated with 0.100 M NaOH, (i) Calculate the volume of OH- required to reach the equivalence point (ii) Calculate the pH when 73 mL of OH- is added: (iii) Calculate the pH when 100 ml of OH- is added: (vi) Calculate the pH when 200 ml of OH- is added: (v) Calculate the pH when 237 ml of OH- is added:

Explanation / Answer

A) Describe the acid-base characteristics of regions B, D and F and points A, C and E.

Buffer regions: B, D and F
Equivalence point: C and E
Initial point: A

(B) Suppose that a titration vessel contains 100.00 ml of 0.100M H2B
(K1 = 1.00 x 10-4 and K2 = 1.00 x 10 –7) and it is titrated with 0.100 M NaOH,

Titration of weak dibasic acid H2B with strng base NaOH
pK1=4, pK2=7

H2B + H2O = H3O+ + HB-

HB- + H2O = H3O+ + B(2-)

pH = 1/2*Log(1+[HB-]/K1)-1/2*Log(Kw+K2*[HB-])

For, [HB-]/K1 >> 1 and K2*[HB-] >> Kw
then, pH = 1/2*Log([HB-]/K1)-1/2*Log(K2*[HB-]) = 1/2*Log(1/(K1*K2)) = 1/2(pK1+pK2)

(i) Calculate the volume of OH- required to reach the equivalence point
For First equivalence point
Equivalent of acid = Molarity of acid*Volume of acid in liter=0.1*0.1=0.01
volume of OH- required = Moles of acid/(Molarity of base)=0.01/0.1=0.1liter=100ml

For second equivalence point
Equivalent of acid = 2*Molarity of acid*Volume of acid in liter=2*0.1*0.1=0.02
volume of OH- required = Moles of acid/(Molarity of base)=0.02/0.1=0.2liter=200ml

(ii) Calculate the pH when 73 mL of OH- is added:
First Buffer region:

pH = pK1 + Log[Salt]/[Acid]

[HB-] = 73*0.1/(73+100)= 4.2*1E-2M
[H2B] = (100*.1-73*.1)/(73+100)=1.5E-2M
pH = 4 + Log[4.2E-2]/[1.5E-2] = 4.44

(iii) Calculate the pH when 100 ml of OH- is added:
First equivalence point(C): pH = (pK1+pK2)/2=(4+7)/2 = 5.5

(iv) Calculate the pH when 200 ml of OH- is added:
Second Equivalence point
[B(2-)]=100*0.1/(100+200)=0.033M
[B(2-)+H2O = OH- + HB-
Kb = Kw/K2=1E-7

(v) Calculate the pH when 237 ml of OH- is added:

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