The solubility product of silver bromide (AgB | vedclass

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

Question

(C) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-]$.Given $[Ag^+] = [AgNO_3] = 0.05 \, M$.Substituting the values into the $K_

Vedclass · Accepted Answer

$1.2 	imes 10^{-9} \, g$

Vedclass · Answer

(C) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-]$.Given $[Ag^+] = [AgNO_3] = 0.05 \, M$.Substituting the values into the $K_{sp}$ expression:$5.0 	imes 10^{-13} = (0.05) 	imes [Br^-]$.$[Br^-] = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \, M$.Since $[Br^-] = [KBr]$,the concentration of $KBr$ required is $1.0 	imes 10^{-11} \, mol \, L^{-1}$.For $1 \, L$ of solution,the number of moles of $KBr$ is $1.0 	imes 10^{-11} \, mol$.Mass of $KBr = 	ext{moles} 	imes 	ext{molar mass} = (1.0 	imes 10^{-11} \, mol) 	imes (120 \, g \, mol^{-1}) = 1.2 	imes 10^{-9} \, g$.

The solubility product of silver bromide $(AgBr)$ is $5.0 \times 10^{-13}$. How much potassium bromide $(KBr)$ (molar mass $120 \, g \, mol^{-1}$) should be added to $1 \, L$ of $0.05 \, M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$?

Similar Questions

The moles of $Ag^{+}$ which must be added to decrease the concentration of $Cl^{-}$ from $4 \times 10^{-5} \ M$ to $10^{-5} \ M$ in $100 \ mL$ solution,if $K_{sp}$ for $AgCl$ is $10^{-10} \ M^2$ at $25 \ ^oC$.

If $K_{sp}$ is the solubility product of $Al(OH)_3$,its solubility is expressed by the formula:

If the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

The $pH$ of a saturated solution of $Ca(OH)_{2}$ is $9$. The solubility product $(K_{sp})$ of $Ca(OH)_{2}$ is:

What is the solubility (in $mol \ L^{-1}$) of $AgCl$ $(K_{sp} = 1.0 \times 10^{-10})$ in a $0.1 \ M \ Cl^-$ solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module