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Question: The solubility product of \(AgBr\) is \(4.9 \times {10^{ - 9}}.\) The solubility of \(AgBr\) will be...

The solubility product of AgBrAgBr is 4.9×10−9.4.9 \times {10^{ - 9}}. The solubility of AgBrAgBr will be.
A. 7×10−4mole/litre7 \times {10^{ - 4}}mole/litre.
B. )7×10−5g/litre7 \times {10^{ - 5}}g/litre.
C. 1.316×10−2g/litre1.316 \times {10^{ - 2}}g/litre.
D. 1×10−3mole/litre1 \times {10^{ - 3}}mole/litre.

Explanation

Solution

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,Ksp=[C]c[D]d{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,
AgBr(s)→Ag +  + Br - {\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 1:11:1 mol ratio.
Given,
The solubility product of silver bromide is 4.9×10−9.4.9 \times {10^{ - 9}}.
Using the solubility product equation we can calculate the solubility of silver bromide as follows,
Ksp=[x][x]{K_{sp}} = \left[ x \right]\left[ x \right]
4.9×10−19=[x][x]4.9 \times {10^{ - 19}} = \left[ x \right]\left[ x \right]
4.9×10−19=x24.9 \times {10^{ - 19}} = {x^2}
x=7×10−4 mol/Litrex = 7 \times {10^{ - 4}}\,mol/Litre
The solubility of silver bromide is7×10−4mole/litre7 \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.