The stoichiometry and solubility product of a | vedclass

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

Question

(C) From the graph,at $[X] = 1 	imes 10^{-3} \ M$,the corresponding $[Y] = 2 	imes 10^{-3} \ M$.Since the concentration of $Y$ is twice that of $X$,

Vedclass · Accepted Answer

$XY_{2}, 4 	imes 10^{-9} \ M^{3}$

Vedclass · Answer

(C) From the graph,at $[X] = 1 	imes 10^{-3} \ M$,the corresponding $[Y] = 2 	imes 10^{-3} \ M$.Since the concentration of $Y$ is twice that of $X$,the stoichiometry of the salt is $XY_{2}$.The dissociation equilibrium is: $XY_{2(s)} ightleftharpoons X_{(aq)}^{2+} + 2Y_{(aq)}^{-}$.The solubility product constant $K_{sp}$ is given by: $K_{sp} = [X^{2+}][Y^{-}]^{2}$.Substituting the values: $K_{sp} = (1 	imes 10^{-3}) 	imes (2 	imes 10^{-3})^{2}$.$K_{sp} = (10^{-3}) 	imes (4 	imes 10^{-6}) = 4 	imes 10^{-9} \ M^{3}$.

The stoichiometry and solubility product of a salt with the solubility curve given below is,respectively:

Similar Questions

Which of the following has the maximum solubility? ($K_{sp}$ values are given in parentheses.)

Which of the following has the minimum solubility in water?

The solubility of $CaCl_2$ is $1.11 \times 10^{-8} \ g/100 \ mL$. If the molar mass of $CaCl_2$ is $111 \ g/mol$,find the $K_{sp}$.

If the solubility product of $BaSO_4$ is $1.44 \times 10^{-12}$,then the solubility of $SO_4^{2-}$ is .....

The solubility of $AgBr$ is $7.1 \times 10^{-7} \ mol \ dm^{-3}$. Calculate its solubility product at the same temperature.

Vedclass Test Series

Exam Paper Generator

Online Exam Module