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If the solubility product of $AgBr{O_3}$ and $A{g_2}S{O_4}$ are $5.5 \times {10^{ - 5}}$ and $2 \times {10^{ - 5}}$ respectively, the relationship between the solubilities of these can be correctly represented as
${S_{AgBr{O_3}}} > {S_{A{g_2}S{O_4}}}$
${S_{AgBr{O_3}}} < {S_{A{g_2}S{O_4}}}$
${S_{AgBr{O_3}}} = {S_{A{g_2}S{O_4}}}$
${S_{AgBr{O_3}}} \approx {S_{A{g_2}S{O_4}}}$
Solution
(b) $A{g_2}S{O_4} \rightleftharpoons \mathop {2A{g^ + }}\limits_{4{S^2}} + \mathop {SO_4^{ – \, – }}\limits_{S\,\,\,\,\,\,\,\,} $
${K_{sp}} = 4{S^3};\,\,{K_{sp}} = 2 \times {10^{ – 5}}$
$S = \sqrt[3]{{\frac{{2 \times {{10}^{ – 5}}}}{4}}} = 0.017\,m/l$$ = 1.7 \times {10^{ – 2}}$
$AgBr{O_3} ⇌\mathop {A{g^ + }}\limits_S + \mathop {BrO_3^ – }\limits_S $
${K_{sp}} = {S^2};\,\,{K_{sp}} = 5.5 \times {10^{ – 5}}$
$S = \sqrt {5.5 \times {{10}^{ – 5}}} = 7.4 \times {10^{ – 3}}\,m/l.$
