The solubility product of AgCl is 1.8 times 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given $K_{sp} = 1.8 	imes 10^{-10}$.When equal v

Vedclass · Accepted Answer

$10^{-4} \ M \ Ag^{+}$ and $10^{-4} \ M \ Cl^{-}$

Vedclass · Answer

(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given $K_{sp} = 1.8 	imes 10^{-10}$.When equal volumes are mixed,the concentration of each ion is halved: $[Ag^{+}]_{new} = [Ag^{+}]_{initial} / 2$ and $[Cl^{-}]_{new} = [Cl^{-}]_{initial} / 2$.For option $A$: $[Ag^{+}] = 10^{-4} / 2 = 5 	imes 10^{-5} \ M$ and $[Cl^{-}] = 10^{-4} / 2 = 5 	imes 10^{-5} \ M$.$Q_{sp} = [Ag^{+}][Cl^{-}] = (5 	imes 10^{-5}) 	imes (5 	imes 10^{-5}) = 25 	imes 10^{-10} = 2.5 	imes 10^{-9}$.Since $2.5 	imes 10^{-9} > 1.8 	imes 10^{-10}$,precipitation will occur.

The solubility product of $AgCl$ is $1.8 \times 10^{-10}$. Mixing equal volumes of which of the following solutions will result in the precipitation of $AgCl$?

Similar Questions

The concentration of a saturated solution of $Mg(OH)_2$ is $8.2 \times 10^{-4} \ g \ L^{-1}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$. (Molar mass of $Mg(OH)_2 = 58 \ g \ mol^{-1}$)

If the solubility product $(K_{sp})$ of a sparingly soluble salt $MX_2$ at $25 \, ^\circ C$ is $1.0 \times 10^{-11}$,what is the solubility of the salt in $mol \, L^{-1}$ at this temperature?

Concentration of the $Ag^{+}$ ions in a saturated solution of $Ag_2C_2O_4$ is $2.2 \times 10^{-4} \ mol \ L^{-1}.$ Solubility product of $Ag_2C_2O_4$ is

The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

Which of the following pairs of ions would be expected to form a precipitate when their dilute solutions are mixed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module