The weak acid, HA has a Ka of 1×`10^(−9)` . If 0.1 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

Options

A. 1%
B. .1%
C. 10%
D. .01%

Answer

(a) (i) `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10CO_(2)+8H_(2)O`
(ii) `Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)to 2Cr^(3+)+7H_(2)O`
(b) (i) On moving from titanium `(Z=22)` to copper `(Z=29)` , electrons get added to the `3d` orbital. So, the mass per unit volume increase increase. As a result, density also increase.
(ii) In the third transition series, due to the introduction of the d orbital, the shielding effect decreases. Therefore, the effective nuclear charge increase. COnsequently, the atomic Volume increases. Due to an increase in mass by volume ratio, density increases.
(iii) The lanthanoids and actinoids exhibit a principal oxidation state of +3. Some lanthanoids also exhibit+4 and +2 oxidation state. However, actinoids exhibit oxidation state of `+2,+4,+5,+6` and `+7`. They exhibit greater range of oxidation state because the `5f,6d` and 7s` sub-shells present in actinoids are of comparable energy . Thus , they can take part in bonding, giving rise to different oxidation states.
OR
(a)(i) `8MnO_(4)^(-)+H_(2)O+3S_(2)O_(3)^(2-) to 8MnO_(2)+6SO_(4)^(2-)+2OH^(-)`
(ii) `Cr_(2)O_(7)^(2-)+3H_(2)S+8H^(+) to 2Cr^(3+)+3S+7H_(2)O`
(b) (i) The gradual decrease in size (actinoid contraction) from element to element is greater among the antinoids than among the lanthanoids. This is because of the poor shielding effect of electrons in the `5f` orbital of acttinoids.
(ii) The elements at the begining of a series exhibit fewer oxidation state because they have less number of electrons which they can lose or contribute for bond formation. The elements at the end of a series exhibit fever oxidation state because they have too many d electrons, and hence , fewer vacant d orbitals that can be involved for bonding.
(iii) For `Cr`, the +3 oxidation state is more stable than the +2 state. Thus, `Cr^(2+)` changes to `Cr^(2+)` and it behaves as a strong reducing agent. However, for Mn, the +2 state is more stable than the +3 state, So,Mn^(3+) changes into `Mn^(2+)` and it behaves as a strong oxidising agent.