Question

The solubility of silver chromate in 0.01 M K2CrO4 is 2*10-8 moldm^3. The solubility product of siver chromate is?

Answer 1

Silver Chromate (Ag2CrO4, 331.73 g/mol) has a Ksp of 1.12 x 10 -12.

a) Calculate the solubility (in g/L) of silver chromate in water.

b) Calculate the solubility (in g/L) of silver chromate in 0.110 M AgNO 3solution.

Answer 2

The solubility product of silver chromate is $3.2\times 10^{-23}$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is represented as $K_{sp}$

Silver chromate is an ionic compound formed by the combination of 2 silver ions and 1 chromate ions

The chemical equation for the ionization of silver chromate follows:

$Ag_2CrO_4(aq.)\rightleftharpoons 2Ag^+(aq.)+CrO_4^{2-}(aq.)$

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=[2Ag^+][CrO_4^{2-}]$

We are given:

$[Ag^+]=(2\times 2\times 10^{-8})=4\times 10^{-8}M$

$[CrO_4^{2-}]=2\times 10^{-8}M$

Putting values in above equation, we get:

$K_{sp}=(4\times 10^{-8})^2\times (2\times 10^{-8})\\\\K_{sp}=3.2\times 10^{-23}$

Learn more about solubility product:

https://brainly.com/question/9515692

https://brainly.in/question/13399466

#learnwithbrainly