JEE MAIN - Chemistry (2010 - No. 4)
Solubility product of silver bromide is 5.0 $$\times$$ 10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the
precipitation of AgBr is :
1.2 $$\times$$ 10–10 g
1.2 $$\times$$ 10–9 g
6.2 $$\times$$ 10–5 g
5.0 $$\times$$ 10–8 g
Explanation
$$AgBr\,\rightleftharpoons\,A{g^ + } + B{r^ - }$$
$${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]$$
For precipitation to occur
Ionic product $$>$$ Solubility product
$$\left[ {B{r^ - }} \right] = {{{K_{sp}}} \over {\left[ {A{g^ + }} \right]}} = {{5 \times {{10}^{ - 13}}} \over {0.05}} = {10^{ - 11}}$$
i.e., precipitation just starts when $${10^{ - 11}}\,$$
moles of $$KBr$$ is added to $$1\ell \,AgN{O_3}\,$$ solution
$$\therefore$$$$\,\,\,$$ Number of moles of $$\,\,\,B{r^ - }\,\,$$ needed
from $$KBr = {10^{ - 11}}$$
$$\therefore$$ $$\,\,\,$$ Mass of $$KBr = {10^{ - 11}} \times 120$$
$$ = 1.2 \times {10^{ - 9}}g$$
$${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]$$
For precipitation to occur
Ionic product $$>$$ Solubility product
$$\left[ {B{r^ - }} \right] = {{{K_{sp}}} \over {\left[ {A{g^ + }} \right]}} = {{5 \times {{10}^{ - 13}}} \over {0.05}} = {10^{ - 11}}$$
i.e., precipitation just starts when $${10^{ - 11}}\,$$
moles of $$KBr$$ is added to $$1\ell \,AgN{O_3}\,$$ solution
$$\therefore$$$$\,\,\,$$ Number of moles of $$\,\,\,B{r^ - }\,\,$$ needed
from $$KBr = {10^{ - 11}}$$
$$\therefore$$ $$\,\,\,$$ Mass of $$KBr = {10^{ - 11}} \times 120$$
$$ = 1.2 \times {10^{ - 9}}g$$
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