Question

At ${\text{25}}{\,^{\text{o}}}{\text{C}}$, ${{\text{K}}_{{\text{sp}}}}$ for ${\text{PbB}}{{\text{r}}_2}$ is$8 \times {10^{ - 5}}$ for. If the salt is $80\% $ dissociated, what is the solubility of ${\text{PbB}}{{\text{r}}_2}$ in ${\text{mol}}\,{\text{/liter}}$?
A. ${\left[ {\dfrac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}} \right]^{\dfrac13}}$
B. ${\left[ {\dfrac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{\dfrac13}}$
C. ${\left[ {\dfrac{{{{10}^{ - 4}}}}{{0.8 \times 0.8}}} \right]^{\dfrac13}}$
D. ${\left[ {\dfrac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}} \right]^{\dfrac12}}$

Answer 1

Hint: ${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as the solubility product.

Step by step answer: The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$
Where,
${\text{AB}}$ is an ionic compound.
The general representation for the solubility product of an ionic compound is as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,{\left( {x{A^{ + y}}\,} \right)^x}\, \times \,\,{\left( {y{A^{ - x}}\,} \right)^y}$
${{\text{K}}_{{\text{sp}}}}\, = \,x{S^x}\, \times \,y{S^y}$
Where,
${{\text{K}}_{{\text{sp}}}}$is the solubility product constant.
${\text{S}}$is the solubility of each ion.
$x$ and $y$ represents the number of ions.
The compound lead bromide dissociates into the water as follows:
${\text{PbB}}{{\text{r}}_2}\, \rightleftarrows \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,2\,{\text{B}}{{\text{r}}^ - }$
Lead bromide in water produces one lead ion and two bromide ions.
The solubility product for chromium hydroxide is represented as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{ + }}\,\,2\,{\text{B}}{{\text{r}}^ - }$
The above solubility represents the $100\% $ dissociation of salt. Here, the salt dissociates only $80\% $so, the solubility of the salt will be $80\% $so,
solubility $\,{\text{ = }}\,\dfrac{{{\text{80}}}}{{{\text{100}}}}\times{\text{ S}}$
solubility $\,{\text{ = }}\,{\text{0}}{\text{.8}}\,{\text{S}}$
Substitute ${\text{0}}{\text{.8}}\,{\text{S}}$for ${\text{S}}$ and $8 \times {10^{ - 5}}$ for ${{\text{K}}_{{\text{sp}}}}$.
\[\Rightarrow 8 \times {10^{ - 5}}\, = \,\left( {{\text{0}}{\text{.8}}\,{\text{S}}\,} \right)\, \times \,\,{\left( {2 \times 0.8\,{\text{S}}} \right)^2}\]
\[\Rightarrow 8 \times {10^{ - 5}} = \,\left( {{\text{0}}{\text{.8}}\,{\text{S}}\,} \right)\, \times \,\,{\left( {1.6\,{\text{S}}} \right)^2}\]
\[\,{{\text{S}}^3}\, = \dfrac{{8 \times {{10}^{ - 5}}}}{{0.8\,\, \times 1.6 \times 1.6}}\]
\[\Rightarrow \,{\text{S}}\, = {\left[ {\dfrac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{\dfrac13}}\]
So, the molar solubility of ${\text{PbB}}{{\text{r}}_2}$in water is ${\left[ {\dfrac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{\dfrac13}}$.
Therefore, option (B) ${\left[ {\dfrac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{\dfrac13}}$, is correct.

Note: The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent.