The solubility product of silver bromide is 5 | vedclass

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

Question

(B) The dissociation of silver bromide is given by: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.For precipitation to start,the ionic product

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The dissociation of silver bromide is given by: $AgBr(s) ightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.For precipitation to start,the ionic product must exceed the solubility product $(K_{sp})$.$K_{sp} = [Ag^{+}][Br^{-}] = 5.0 	imes 10^{-13}$.Given $[Ag^{+}] = 0.05 \ M$,the concentration of $Br^{-}$ required is:$[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \ M$.Since the volume of the solution is $1 \ L$,the number of moles of $Br^{-}$ (or $KBr$) required is $1.0 	imes 10^{-11} \ mol$.The mass of $KBr$ required is: $	ext{Mass} = 	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 is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \ g \ mol^{-1}$) to be added to $1 \ L$ of $0.05 \ M$ solution of silver nitrate to start the precipitation of $AgBr$ is

Similar Questions

When equal volumes of the following solutions are mixed,precipitation of $AgCl$ $(K_{sp} = 1.8 \times 10^{-10})$ will occur only with:

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. Then the $K_{sp}$ for $BaSO_4$ will be:

The solubility product $(K_{sp})$ for sodium chloride is $36 \text{ mol}^2/\text{L}^2$. The solubility of sodium chloride is:

The $K_{sp}$ expression for $Ca_3(PO_4)_2$ is:

Calculate the molar solubility of calcium hydroxide $Ca(OH)_2$ in $0.10 \ M \ NaOH$ solution. The solubility product constant $(K_{sp})$ of calcium hydroxide is $5.5 \times 10^{-6}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module