Question

The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which statement is true?

a. Silver chloride is completely soluble in water.
b. Silver chloride is essentially insoluble in water.
c. No right choice.

Answer 1

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

Step-by-step explanation:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant
`\left(\boldsymbol{K}_(s p)\right)` for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate
`\left(\boldsymbol{K}_(s p)\right)` firstly molarity of ions are needed to be found with formula:
`\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}`

Then at equilibrium cations and anions concentration is considered same hence:

`\left[\mathbf{A} \mathbf{g}^(+)\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }`

Hence from above data
`\left(\boldsymbol{K}_(s p)\right)` can be calculated by:
`\left(\boldsymbol{K}_(s p)\right)` =
`\left[\mathbf{A} \mathbf{g}^(+)\right] \cdot[\mathbf{C} \mathbf{I}]`