The solubility product of Ag_{2}CrO_{4} is 32 | vedclass

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

Question

(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} \rightleftharpoons 2 Ag^{+} + CrO_{4}^{2-}$.Let the solubility be $s$. Then $[Ag^{

Vedclass · Accepted Answer

$2 	imes 10^{-4} \ M$

Vedclass · Answer

(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} ightleftharpoons 2 Ag^{+} + CrO_{4}^{2-}$.Let the solubility be $s$. Then $[Ag^{+}] = 2s$ and $[CrO_{4}^{2-}] = s$.The solubility product expression is $K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2-}]$.Substituting the values: $K_{sp} = (2s)^{2} (s) = 4s^{3}$.Given $K_{sp} = 32 	imes 10^{-12}$,we have $4s^{3} = 32 	imes 10^{-12}$.$s^{3} = 8 	imes 10^{-12}$.$s = (8 	imes 10^{-12})^{1/3} = 2 	imes 10^{-4} \ M$.Since $[CrO_{4}^{2-}] = s$,the concentration is $2 	imes 10^{-4} \ M$.

The solubility product of $Ag_{2}CrO_{4}$ is $32 \times 10^{-12}$. What is the concentration of $CrO_{4}^{2-}$ ions in that solution?

Similar Questions

Write the formula for the solubility of a salt of type $AB_2$.

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 $25 \ ^oC$.

The low solubility of $BaSO_4$ in water can be attributed to

Which of the following is most soluble in water?

The solubility of $Ba_3(AsO_4)_2$ (molar mass $= 690 \ g/mol$) is $6.9 \times 10^{-2} \ g/100 \ mL$. What is the value of its $K_{sp}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module