Question

The solubility product of \[{\text{AgCl}}\] is \[1.8 \times {10^{ - 10}}\]. Precipitation of \[{\text{AgCl}}\] will occur only when equal volumes of solution of:
A.\[{10^{ - 4}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 4}}{\text{M C}}{{\text{l}}^ - }\] are mixed.
B.\[{10^{ - 7}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 7}}{\text{M C}}{{\text{l}}^ - }\] are mixed
C.\[{10^{ - 5}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 5}}{\text{M C}}{{\text{l}}^ - }\] are mixed
D.\[2 \times {10^{ - 5}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 5}}{\text{M C}}{{\text{l}}^ - }\] are mixed

Answer 1

Hint: Precipitation will only occur when the ionic product of the two ions in silver chloride will be greater than the solubility product. Check the ionic product in each of the options and the one with a greater product than the given solubility product will be the correct answer.

Complete step by step answer:
Let us calculate the ionic product for each of the options; if it is greater than that of solubility product then the formation of precipitate will occur. Silver chloride dissociated as:
\[{\text{AgCl}} \to {\text{A}}{{\text{g}}^ + } + {\text{ C}}{{\text{l}}^ - }\]
When a mixture of \[{10^{ - 4}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 4}}{\text{M C}}{{\text{l}}^ - }\] is there then we will get ionic product:
\[{{\text{K}}_{\text{I}}} = {\text{ [A}}{{\text{g}}^ + }{\text{] [C}}{{\text{l}}^ - }]\]
\[{{\text{K}}_{\text{I}}} = {\text{ }}{10^{ - 4}} \times {10^{ - 4}} = {10^{ - 8}}\]
We know \[{10^{ - 8}} > 1.8 \times {10^{ - 10}}\] the value of ionic products comes out greater than solubility products. Hence the precipitate will form in this. Since power is negative, the greater value will be actually smaller.
For a mixture of \[{10^{ - 7}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 7}}{\text{M C}}{{\text{l}}^ - }\] is there then we will get ionic product:
\[{{\text{K}}_{\text{I}}} = {\text{ [A}}{{\text{g}}^ + }{\text{] [C}}{{\text{l}}^ - }]\]
\[{{\text{K}}_{\text{I}}} = {\text{ }}{10^{ - 7}} \times {10^{ - 7}} = {10^{ - 14}}\]
Here, \[{10^{ - 14}} < 1.8 \times {10^{ - 10}}\] no precipitate will form.
For a mixture of \[{10^{ - 5}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 5}}{\text{M C}}{{\text{l}}^ - }\] is there then we will get ionic product:
\[{{\text{K}}_{\text{I}}} = {\text{ [A}}{{\text{g}}^ + }{\text{] [C}}{{\text{l}}^ - }]\]
\[{{\text{K}}_{\text{I}}} = {\text{ }}{10^{ - 5}} \times {10^{ - 5}} = {10^{ - 10}}\]
Here, \[{10^{ - 10}} = {10^{ - 10}}\] no precipitate will form.
For a mixture of \[2 \times {10^{ - 5}}{\text{M A}}{{\text{g}}^ + }{\text{ and }}{10^{ - 5}}{\text{M C}}{{\text{l}}^ - }\] is there then we will get ionic product:
\[{{\text{K}}_{\text{I}}} = {\text{ [A}}{{\text{g}}^ + }{\text{] [C}}{{\text{l}}^ - }]\]
\[{{\text{K}}_{\text{I}}} = {\text{ 2}} \times {10^{ - 5}} \times {10^{ - 5}} = 2 \times {10^{ - 10}}\]
Here, \[2 \times {10^{ - 10}} \simeq 1.8 \times {10^{ - 10}}\]. They are approximately similar and hence, no precipitate will form.

Hence, the correct option is A.

Note:
The term solubility product is defined for only sparingly soluble salt. Sparingly soluble salts are those salts which dissociate partially. Greater is the Solubility constant for a salt, greater will be its solubility in aqueous medium.