The required amount of KBr (molar mass = 119 | vedclass

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

Question

(C) Precipitation of $AgBr$ starts when the ionic product exceeds the solubility product $(K_{sp})$.$AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$$K

Vedclass · Accepted Answer

$5.95 	imes 10^{-10} \, g$

Vedclass · Answer

(C) Precipitation of $AgBr$ starts when the ionic product exceeds the solubility product $(K_{sp})$.$AgBr(s) ightleftharpoons Ag^+(aq) + Br^-(aq)$$K_{sp} = [Ag^+][Br^-]$Given $[Ag^+] = 0.05 \, M$ and $K_{sp} = 5 	imes 10^{-13}$.$[Br^-] = \frac{K_{sp}}{[Ag^+]} = \frac{5 	imes 10^{-13}}{0.05} = 10^{-11} \, M$.Number of moles of $Br^-$ required $= Molarity 	imes Volume(L) = 10^{-11} \, mol/L 	imes 0.5 \, L = 0.5 	imes 10^{-11} \, mol$.Since $KBr$ dissociates as $KBr ightarrow K^+ + Br^-$,moles of $KBr$ required $= 0.5 	imes 10^{-11} \, mol$.Mass of $KBr = moles 	imes molar \, mass = 0.5 	imes 10^{-11} 	imes 119 = 59.5 	imes 10^{-11} = 5.95 	imes 10^{-10} \, g$.

The required amount of $KBr$ (molar mass $= 119 \, g/mol$) in grams to start the precipitation of $AgBr$ in $500 \, mL$ solution of $0.05 \, M \, AgNO_3$ will be ($K_{sp}$ of $AgBr = 5 \times 10^{-13}$)

Similar Questions

If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$,then the equation of the circle is

If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{3} = 4$ is $\frac{\pi}{3}$,then its conjugate hyperbola is:

Identify the plant shown in the given figure.

Which one of the following does not correctly represent the order of the property indicated against it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module