JEE Advance - Chemistry (2018 - Paper 1 Offline - No. 9)
The solubility of a salt of weak acid $$(AB)$$ at $$pH\,$$ $$3$$ is $$Y \times {10^{ - 3}}$$ $$mol\,{L^{ - 1}}.$$ The value of $$Y$$ is ________________.
(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)
(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)
Answer
4.47
Explanation
Let solubility of AB in the buffer of pH 3 = x
K3 = $${{{K_{sp}}} \over {{K_a}}}$$
$${K_3} = {{[HB][{A^ + }]} \over {[{H^ + }]}} = {{{K_{sp}}} \over {{K_a}}}$$
$$\therefore$$ $${{{x^2}} \over {({{10}^{ - 3}})}} = {{2 \times {{10}^{ - 10}}} \over {1 \times {{10}^{ - 8}}}}$$
$$\therefore$$ x = 4.47 $$\times$$ 10$$-$$3 M = y $$\times$$ 10$$-$$3 M
$$\therefore$$ y = 4.47
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