Question

The molar solubility of $ {\text{PbB}}{{\text{r}}_2} $ is $ 2.17 \times {10^{ - 3}}{\text{M}} $ at a certain temperature. Calculate $ {{\text{K}}_{{\text{sp}}}} $ for $ {\text{PbB}}{{\text{r}}_2} $ .

Answer 1

Hint: The ability of a substance called a solute to dissolve in a solvent and form a solution is known as solubility. Ionic compounds (which dissociate to form cations and anions) have a wide range of solubility in water. Some compounds are very soluble and can also absorb moisture from the air, while others are extremely insoluble.

Complete answer:
When a chemical compound in the solid state is in chemical equilibrium with a solution of that compound, it is called solubility equilibrium. The solid can dissolve in its entirety, through dissociation, or through chemical reaction with another solution component, such as acid or alkali. A temperature-dependent solubility substance, which acts as an equilibrium constant, characterises each solubility equilibrium. Solubility equilibria are essential in a variety of situations, including pharmaceuticals, the climate, and many others.
 $ K = \dfrac{{\left[ {P{b^{2 + }}} \right]{{\left[ {B{r^ - }} \right]}^2}}}{{\left[ {PbB{r_2}} \right]}} $
The equilibrium constant for the dissolution of a solid material into an aqueous solution is the solubility product constant. The symbol $ {{\text{K}}_{{\text{sp}}}} $ is used to represent it. The solubility product is a type of equilibrium constant whose value is temperature dependent. Due to enhanced solubility, $ {{\text{K}}_{{\text{sp}}}} $ typically rises as the temperature rises.
The molar solubility of $ {\text{PbB}}{{\text{r}}_2} $ is $ 2.17 \times {10^{ - 3}}{\text{M}} $
 $ \begin{array}{*{20}{l}}
  {{\text{PbB}}{{\text{r}}_2} \rightleftharpoons {\text{P}}{{\text{b}}^{ + 2}} + 2{\text{B}}{{\text{r}}^ - }} \\
  {{\text{s s 2s}}}
\end{array} $
Using $ {{\text{K}}_{{\text{sp}}}} $ formula
 $ {K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b} $
 $ {{\text{K}}_{{\text{sp}}}} = \left[ {{\text{P}}{{\text{b}}^{ + 2}}} \right]{\left[ {{\text{2B}}{{\text{r}}^ - }} \right]^2} $
 $ {{\text{K}}_{{\text{SP}}}} = \left[ {{\mathbf{P}}{{\text{b}}^{2 + }}} \right]{\left[ {{\text{2B}}{{\text{r}}^ - }} \right]^2} = ({\text{S}}){(2\;{\text{S}})^2} = 4\;{{\text{S}}^3} $
Substitute S = $ 2.17 \times {10^{ - 3}}{\text{M}} $
 $ {{\text{K}}_{{\text{SP}}}} = 4\;{{\text{S}}^3} = 4{\left( {2.17 \times {{10}^{ - 3}}} \right)^3} = 4.1 \times {10^{ - 8}} $
Solubility equilibria can be divided into three categories.
Simple dissolution.
Dissociation reaction for dissolution. This is a common feature of salts. In this case, the equilibrium constant is referred to as a solubility product.
Ionization reaction for dissolution. This is typical of weak acids or weak bases dissolving in aqueous media with differing pH.

Note:
An equilibrium constant can be defined as a quotient of activities in each case. Since action is a dimensionless quantity, this equilibrium constant is dimensionless as well. However, since using behaviours is inconvenient, the equilibrium constant is generally separated by the quotient of activity coefficients, resulting in a concentration quotient.