The solubility product of silver bromide (AgB | vedclass

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

Question

(B) The precipitation of $AgBr$ begins when the ionic product exceeds the solubility product $(K_{sp})$.$K_{sp} = [Ag^+][Br^-] = 5.0 	imes 10^{-13}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The precipitation of $AgBr$ begins when the ionic product exceeds the solubility product $(K_{sp})$.$K_{sp} = [Ag^+][Br^-] = 5.0 	imes 10^{-13}$.Given $[Ag^+] = 0.05 \, M$,we find the concentration of $Br^-$ required:$[Br^-] = \frac{K_{sp}}{[Ag^+]} = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \, M$.Since the volume is $1 \, L$,the number of moles of $KBr$ required 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 $5.0 \times 10^{-13}$. Calculate the mass of potassium bromide $(KBr)$ (molar mass = $120 \, g \, mol^{-1}$) that must be added to $1 \, L$ of a $0.05 \, M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$.

Similar Questions

What is the coefficient of mutual inductance when the magnetic flux changes by $2 \times 10^{-2} \, Wb$ and the change in current is $0.01 \, A$?

In a $\triangle ABC$,$a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

$\sqrt[4]{17 + 12\sqrt{2}} = $

Laser beams are used to measure long distances because

Vedclass Test Series

Exam Paper Generator

Online Exam Module