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Answer:

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

Data Given:

              Ksp of CaSO4 = 2.4 × 10∧-5

              CaSO4 ⇔ Ca+2   +  SO4∧-2

Solution:

                Ksp = [Ca+2].[ SO4∧-2]

                 2.4 × 10∧-5 = [x].[x]= x²

                 x =  4.9×10∧-3 M

Result:

  • The conc. of Ca+2 ion is 4.9×10∧-3 M

The study of chemicals and bonds is called chemistry. When the amount of product and the reactant is equal then is said to the equilibrium.

The correct answer is [tex]4.9*10^{-3}[/tex] M

What is equilibrium constant?

  • The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
  • A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change

The Data is Given as follow

  • Ksp of[tex]CaSO_4 = 2.4 * 10{^-5}[/tex]
  • [tex]CaSO_4 <---> Ca^{+2} + SO_4^{-2}[/tex]

The solution is as follows:-

[tex]Ksp = [Ca^{+2}].[ SO_4^{-2}]\\\\2.4 * 10^{-5} \\= [x].[x]= x^2\\ x = 4.9810^{-3}M[/tex]

After solving the equation, the concentration will be[tex]4.9*10^{-3}M[/tex]

Hence, the correct answer is [tex]4.9*10^{-3}M[/tex].

For more information about the equilibrium, refer to the link:-

https://brainly.com/question/787658

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