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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
A 0.02 M solution of a weak mono basic acid is 5% ionised. Calculate the ionisation constant of the acid.
2.
The degree of dissociation of acetic acid in 0.1 M solution is 0.04. Calculate Ka for acetic acid. Where a is the degree of dissociation, C is the concentration of the acid in moles/ lit.
3.
If a solution has a pH of 7.41, determine its H+ concentration.
4.
Calculate the pH of 0.001 M HCI solution.
5.
Calculate the pH of 0.02 m Ba(OH)2 aqueous solution assuming Ba(OH)2 as a strong electrolyte.
1.
The degree of ionisation and the dissociation constant of the weak acid are related by the equation.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } \cong { \alpha }^{ 2 }C\)
α = 5% (or) 0.05
C = 0.02M
Ka = (0.05)2 x 0.02 = 0.00005
Ka = 5 x 10-5
2.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =\frac { .02\times .02\times 0.1 }{ 1-0.02 } \)
\({ K }_{ a }=\frac { 0.4\times { 10 }^{ -4 } }{ 0.98 } =4.08\times { 10 }^{ -5 }\)
3.
pH = -log [H+]
∴ [H+] = antilog [-pH]
= antilog [-7.41]
∴ [H+] = 3.9 x 10-8 M.
4.
HCI ⟶ H+ + Cl-. HCI is a strong acid.
[H+] from HCI is very much greater than [H+] from water which is 1 x 10-7 M.
∴ [H+] = [HCI] = 0.001 M
∴ pH = -log (0.001) = 3.0
∴ That is acidic solution.
5.
Ba(OH2) ⟶ Ba2+ 2OH-
∴ [OH-] = 2 [Ba(OH)2]
= 2 x 0.02 = 0.04 M
∴ pOH = -log [OH-]
= 1.398 = 1.40
∴ pH = 14 - 1.4 = 12.6
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