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Question: Solubility product of silver bromide is \(5.0 \times {10^{ - 13}}\) . The quantity of potassium brom...

Solubility product of silver bromide is 5.0×10−135.0 \times {10^{ - 13}} . The quantity of potassium bromide (molar mass taken as 120gmol−1120gmo{l^{ - 1}} ) to be added to one litre of 0.05M0.05M solution of silver nitrate to start the precipitate of AgBrAgBr is:
A. 1.2×10−10g1.2 \times {10^{ - 10}}g
B. 1.2×10−9g1.2 \times {10^{ - 9}}g
C. 6.2×10−5g6.2 \times {10^{ - 5}}g
D. 5.0×10−8g5.0 \times {10^{ - 8}}g

Explanation

Solution

Hint: Solubility product is denoted by Ksp{K_{sp}} which is called the solubility product constant. The smaller the solubility product, the lower the solubility. Solubility product tells whether a precipitate will form under specified conditions, or not. Solubility products of any substance is the ratio of concentrations at equilibrium. Molar solubility is related to the solubility product.

Complete answer:
In the problem it is given that the concentration of silver bromide, concentration of silver ion, [Ag+]=0.05M[A{g^ + }] = 0.05M, AgBr⇌Ag++Br−AgBr \rightleftharpoons A{g^ + } + B{r^ - }
⇒Ksp=[Ag+][Br−]\Rightarrow {K_{sp}} = [A{g^ + }][B{r^ - }] ;
⇒[Br−]=Ksp[Ag+]\Rightarrow [B{r^ - }] = \dfrac{{{K_{sp}}}}{{[A{g^ + }]}};
Now we know that solubility product of silver bromide is 5.0×10−135.0 \times {10^{ - 13}} means [Ag+][Br−]=5×10−13[A{g^ + }][B{r^ - }] = 5 \times {10^{ - 13}}.
⇒[Br−]×V=10−11molL−1×1L\Rightarrow [B{r^ - }] \times V = {10^{ - 11}}mol{L^{ - 1}} \times 1L
⇒[Br−]=5×10−130.05\Rightarrow [B{r^ - }] = \dfrac{{5 \times {{10}^{ - 13}}}}{{0.05}} =10−11M = {10^{ - 11}}M , it means that there is 10−11{10^{ - 11}} moles of potassium bromide in on litre of solution i.e. on addition of 10−11{10^{ - 11}} moles of KBrKBr in one litre of solution precipitation will start.
Molar mass of KBr=120gmol−1KBr = 120gmo{l^{ - 1}}(given in the problem) ,the quantity of potassium bromide to be added to one litre of 0.05M0.05M solution of silver nitrate to start the precipitate of AgBrAgBr is ;10−11mol×120gmol−1=1.2×10−9g{10^{ - 11}}mol \times 120gmo{l^{ - 1}} = 1.2 \times {10^{ - 9}}g .

Note: We have approached this problem by using the concept of solubility product. Here in the problem solubility product of silver bromide is given which is equal to 5.0×10−135.0 \times {10^{ - 13}} .With the help of solubility product we find out concentration of bromide ion or we can say that we find moles of potassium bromide in the solution and the molar mass of potassium bromide is given .On multiplying molar mass of potassium bromide with the concentration of bromine ion we get the answer which is 1.2×10−9g1.2 \times {10^{ - 9}}g . So the quantity of potassium bromide to be added to one litre of 0.05M0.05M solution of silver nitrate to start the precipitate of AgBrAgBr is 1.2×10−9g1.2 \times {10^{ - 9}}g.