Question

The solubility product of $AgBr$ is $4.9 \times {10^{ - 9}}.$ The solubility of $AgBr$ will be.
A. $7 \times {10^{ - 4}}mole/litre$.
B. )$7 \times {10^{ - 5}}g/litre$.
C. $1.316 \times {10^{ - 2}}g/litre$.
D. $1 \times {10^{ - 3}}mole/litre$.

Answer 1

Hint: Now, we discuss the solubility product.
The equilibrium constant of a solid substance dissolved in an aqueous solution is termed as the solubility product.
The mathematical expression is given as,${K_{sp}} = {\left[ C \right]^c}{\left[ D \right]^d}$
To calculate the solubility product the concentration of each ion in molarity is needed.

Complete step by step answer: Consider the dissociation reaction,
${\text{AgBr}}\left( s \right)\xrightarrow{{}}{\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + B}}{{\text{r}}^{\text{ - }}}$
If the solubility of silver bromide is x then the solubility of each product ion would also be $1:1$ mol ratio.
Given,
The solubility product of silver bromide is $4.9 \times {10^{ - 9}}.$
Using the solubility product equation we can calculate the solubility of silver bromide as follows,
${K_{sp}} = \left[ x \right]\left[ x \right]$
$4.9 \times {10^{ - 19}} = \left[ x \right]\left[ x \right]$
$4.9 \times {10^{ - 19}} = {x^2}$
$x = 7 \times {10^{ - 4}}\,mol/Litre$
The solubility of silver bromide is$7 \times {10^{ - 4}}mole/litre$.
So, the correct answer is “Option A”.

Note: Let we see the details about solubility product
Solubility Product:
- We must know that the equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution is known as "solubility product". It is denoted by ''.
- Solubility products are also known as "ion products".
- The value of solubility products usually increases with an increase in temperature due to increased solubility.
- Some of the factors that affect the value of solubility products are the common-ion effect, the diverse-ion effect, and the presence of ion-pairs.
- If the ionic concentrations have less value than the solubility product, the solution isn't saturated. No precipitate would be formed.
- If the ionic concentration has more value than the solubility product, sufficient precipitate would be formed to decrease concentrations to answer similar to the solubility product.