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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
Calculate the pH of 0.001M HCl solution
2.
Calculate the concentration of OH- in a fruit juice which contains \(2\times10^{-3}\) M, H3O+ ion. Identify the nature of the solution.
3.
What are Lewis acids and bases? Give two example for each.
4.
Discuss the Lowry – Bronsted concept of acids and bases.
5.
Calculate pH of 10-7 M HCl
1.
\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H3O+ from the auto ionisation of H2O (10-7M) is negligible when compared to the H3O+ from 10-3M HCl.
Hence [H3O+] = 0.001 mol dm-3
pH = -log10 [H3O+]
= -log10(0.001)
= -log10(10-3) = 3
2.
Given that H3O+ = \(2\times10^{-3}M\)
\(K_{w}=[H_{3}O^{+}][OH^{-}]\)
\(\therefore [OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1\times10^{-14}}{2\times10^{-3}}=0.5\times10^{-11}M\)
\(2\times10^{-3} >>0.5\times10^{-11}\)
i.e., [H3O+]>>[OH-], hence the juice is acidic in nature
3.
(i) Lewis acid: It is a species that accepts an electron pair. Eg: \(\mathrm{Ag}^{+} ; \mathrm{BF}_{3} ; \mathrm{A} / \mathrm{Cl}_{3}\)
(ii) Lewis base: It is a species that donates an electron pair. Eg: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)
4.
(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H2O is base. The proton transfer from the acid to base can be represented as
HCI + H2O ⇌ H3O+ + Cl-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH3) acts as a base and H2O is acid. The reaction is represented as
H2O + NH3 ⇌ NH4+ + OH-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \)
H3O+ donates a proton to Cl- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as
Acid1 + Base2 ⇌ Acid2 + Base1
(vii) The species that remains after the donation of a proton is a base (Base1) and is called the conjugate base of the Bronsted acid (Acid1). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.
5.
If we do not consider [H3O]+ from the ionisation of H2O,
then [H3O+] = [HCl] = 10-7M
i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10-7M) Hence, the H3O+ (10-7M) formed due to the auto ionisation of water cannot be neglected.
so, in this case we should consider [H3O+] from ionisation of H2O
[H3O+] = 10-7 (from HCl) + 10-7 (from water)
= 10-7 (1+1)
= \(2\times10^{-7}\)
pH = -log10[H3O+]
=\(-\log_{10}(2\times10^{-7})=-[\log2+\log_{10}10^{-7}]\)
=\(-\log2-(-7)\log_{10}^{10}\)
= 7-log2
= 7-0.3010 = 0.6990 = 6.70
= 6.70
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