Calculate simultaneous solubility of $AgCNS$ and $AgBr$ in a solution of water.
$[Given:{{K}_{sp}}ofAgCNS=1\times {{10}^{-12}},{{K}_{sp}} ofAgBr=5\times {{10}^{-13}}]$
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Hint: Here, see $AgCNS$ and $AgBr$ have a COMMON ION within them. Calculate the solubility with the ${{K}_{sp}}$ formula i.e. ${{K}_{sp}}$ = ${{\left[ A \right]}^{p}}{{\left[ B \right]}^{q}}$ (product of the equilibrium concentrations of the ions in a saturated solution of a salt.)
Complete answer:
Solubility is the property of a solid, liquid or gaseous chemical substance i.e. solute to get dissolved in another solid, liquid or gaseous solvent. Temperature, pressure, and presence of other chemicals ( including changes to the $pH$ ) of the solution are the fundamental physical and chemical factors upon which solubility of a substance depends on.
By the method of saturation concentration, the extent of the solubility of a specific substance is measured. The point at which adding more solute does not increase the concentration of the solution but begins to produce a precipitate we assume that saturation concentration has been reached.
Now, we should also know about the concept of Solubility product (${{K}_{sp}}$).
Sparingly soluble salts do not dissociate completely into ions and therefore, their ionic product can’t be calculated .Thus, the term Solubility product is applicable for these salts. It is the maximum product of the molar concentration of the ions which are produced due to the dissociation of the compound.
For a chemical equilibrium,
${{A}_{p}}{{B}_{q}}\rightleftharpoons pA+qB$
The Solubility product, ${{K}_{sp}}$ for the compound ${{A}_{p}}{{B}_{q}}$ is defined as follows,
${{K}_{sp}}={{[A]}^{p}}{{[B]}^{q}}$
where$[A]$ and $[B]$ are the concentrations of A and B in a saturated solution.
The solubility product is just like equilibrium constant and has the dimension of${{(concentration)}^{p+q}}$ .
Now, let’s do the calculation,
Let the solubility of $AgCNS$ be a and that of$AgBr$ be b.
∴ $AgCNS\rightleftharpoons A{{g}^{+}}+CN{{S}^{-}}$
At equilibrium , a a
\[AgBr\rightleftharpoons A{{g}^{+}}+\text{ }B{{r}^{}}\]
At equilibrium, b b
∴ ${{K}_{sp}}$ of$AgCNS$ =\[\left[ A{{g}^{+}} \right]\text{ }\left[ CN{{S}^{-}} \right]\text{ }=~\left( a\text{ }+\text{ }b \right)\text{ }a\text{ }=\text{ }{{10}^{-12}}\]
Similarly, ${{K}_{sp}}$ of = \[\left[ A{{g}^{+}} \right]\text{ }\left[ B{{r}^{-}} \right]\] = (a + b) b = $5\text{ }\times {{10}^{(-13)}}.$
Now, by dividing both the ${{K}_{sp}}$ .values, we get,
$\therefore \tfrac{{{K}_{sp}}(AgCNS)}{{{K}_{sp}}(AgBr)}=\tfrac{{{10}^{-12}}}{5\times {{10}^{-13}}}=\tfrac{(a+b)a}{(a+b)b}$
$6{{b}^{2}}={{10}^{(-12)}}$ $\Rightarrow \tfrac{a}{b}=\tfrac{{{10}^{13}}}{5\times {{10}^{12}}}=\tfrac{10}{5}=2$
$\Rightarrow a=2b$ . $-Equation\left( 1 \right)$
Now, $a(a+b)={{10}^{-12}}$
Using Equation (1),
∴ $2b(2b+b)={{10}^{(-12)}}$
$\Rightarrow $ $4{{b}^{2}}+2{{b}^{2}}={{10}^{(-12)}}$
$\Rightarrow $ $6{{b}^{2}}={{10}^{(-12)}}$
$\Rightarrow {{b}^{2}}=17\times {{10}^{-12}}$
$\Rightarrow b=4.123\text{ }\times {{10}^{(-6)}}.$ $-Equation\left( 2 \right)$
Similarly,$(a+b)a={{10}^{-12}}$
$(a+b)a={{10}^{-12}}$
Putting the values of ‘b’ from Equation 2,
$\Rightarrow (a+4.123\text{ }\times {{10}^{(-6)}})4.123\text{ }\times {{10}^{(-6)}}={{10}^{(-12)}}$
$\Rightarrow (4.123\text{ }\times {{10}^{(-6)}})a+17\text{ }\times {{10}^{(-12)}}={{10}^{(-12)}}$
$\Rightarrow (4.123\text{ }\times {{10}^{(-6)}})a=-1.6\times {{10}^{(-11)}}$
$\Rightarrow a=-3.9\text{ }\times {{10}^{(-18)}}$
Here, the Common Ion effect is seen. $A{{g}^{+}}$ ion is present in both $AgCNS$ and $AgBr$ . Decrease in solubility of a salt is observed when it is dissolved in a solution that already contains different sources of its ions present in it.
The simultaneous solubility of $AgCNS$ and $AgBr$ in a solution of water are $a=-3.9\times {{10}^{-18}}$ and $b=4.123\times \text{ }{{10}^{(-6)}}$ respectively.
Note: Do the ${{K}_{sp}}$ calculations carefully and see if there is a common ion or not. The common ion like here it is for $A{{g}^{+}}$has an effect on the solubility of the solution.
Complete answer:
Solubility is the property of a solid, liquid or gaseous chemical substance i.e. solute to get dissolved in another solid, liquid or gaseous solvent. Temperature, pressure, and presence of other chemicals ( including changes to the $pH$ ) of the solution are the fundamental physical and chemical factors upon which solubility of a substance depends on.
By the method of saturation concentration, the extent of the solubility of a specific substance is measured. The point at which adding more solute does not increase the concentration of the solution but begins to produce a precipitate we assume that saturation concentration has been reached.
Now, we should also know about the concept of Solubility product (${{K}_{sp}}$).
Sparingly soluble salts do not dissociate completely into ions and therefore, their ionic product can’t be calculated .Thus, the term Solubility product is applicable for these salts. It is the maximum product of the molar concentration of the ions which are produced due to the dissociation of the compound.
For a chemical equilibrium,
${{A}_{p}}{{B}_{q}}\rightleftharpoons pA+qB$
The Solubility product, ${{K}_{sp}}$ for the compound ${{A}_{p}}{{B}_{q}}$ is defined as follows,
${{K}_{sp}}={{[A]}^{p}}{{[B]}^{q}}$
where$[A]$ and $[B]$ are the concentrations of A and B in a saturated solution.
The solubility product is just like equilibrium constant and has the dimension of${{(concentration)}^{p+q}}$ .
Now, let’s do the calculation,
Let the solubility of $AgCNS$ be a and that of$AgBr$ be b.
∴ $AgCNS\rightleftharpoons A{{g}^{+}}+CN{{S}^{-}}$
At equilibrium , a a
\[AgBr\rightleftharpoons A{{g}^{+}}+\text{ }B{{r}^{}}\]
At equilibrium, b b
∴ ${{K}_{sp}}$ of$AgCNS$ =\[\left[ A{{g}^{+}} \right]\text{ }\left[ CN{{S}^{-}} \right]\text{ }=~\left( a\text{ }+\text{ }b \right)\text{ }a\text{ }=\text{ }{{10}^{-12}}\]
Similarly, ${{K}_{sp}}$ of = \[\left[ A{{g}^{+}} \right]\text{ }\left[ B{{r}^{-}} \right]\] = (a + b) b = $5\text{ }\times {{10}^{(-13)}}.$
Now, by dividing both the ${{K}_{sp}}$ .values, we get,
$\therefore \tfrac{{{K}_{sp}}(AgCNS)}{{{K}_{sp}}(AgBr)}=\tfrac{{{10}^{-12}}}{5\times {{10}^{-13}}}=\tfrac{(a+b)a}{(a+b)b}$
$6{{b}^{2}}={{10}^{(-12)}}$ $\Rightarrow \tfrac{a}{b}=\tfrac{{{10}^{13}}}{5\times {{10}^{12}}}=\tfrac{10}{5}=2$
$\Rightarrow a=2b$ . $-Equation\left( 1 \right)$
Now, $a(a+b)={{10}^{-12}}$
Using Equation (1),
∴ $2b(2b+b)={{10}^{(-12)}}$
$\Rightarrow $ $4{{b}^{2}}+2{{b}^{2}}={{10}^{(-12)}}$
$\Rightarrow $ $6{{b}^{2}}={{10}^{(-12)}}$
$\Rightarrow {{b}^{2}}=17\times {{10}^{-12}}$
$\Rightarrow b=4.123\text{ }\times {{10}^{(-6)}}.$ $-Equation\left( 2 \right)$
Similarly,$(a+b)a={{10}^{-12}}$
$(a+b)a={{10}^{-12}}$
Putting the values of ‘b’ from Equation 2,
$\Rightarrow (a+4.123\text{ }\times {{10}^{(-6)}})4.123\text{ }\times {{10}^{(-6)}}={{10}^{(-12)}}$
$\Rightarrow (4.123\text{ }\times {{10}^{(-6)}})a+17\text{ }\times {{10}^{(-12)}}={{10}^{(-12)}}$
$\Rightarrow (4.123\text{ }\times {{10}^{(-6)}})a=-1.6\times {{10}^{(-11)}}$
$\Rightarrow a=-3.9\text{ }\times {{10}^{(-18)}}$
Here, the Common Ion effect is seen. $A{{g}^{+}}$ ion is present in both $AgCNS$ and $AgBr$ . Decrease in solubility of a salt is observed when it is dissolved in a solution that already contains different sources of its ions present in it.
The simultaneous solubility of $AgCNS$ and $AgBr$ in a solution of water are $a=-3.9\times {{10}^{-18}}$ and $b=4.123\times \text{ }{{10}^{(-6)}}$ respectively.
Note: Do the ${{K}_{sp}}$ calculations carefully and see if there is a common ion or not. The common ion like here it is for $A{{g}^{+}}$has an effect on the solubility of the solution.