Question

Course Home <Chapter 17 part 2 Exercise 17.105 Part C Review | Constants l Periodic Table Consider a solution that is 1.5x10-2 M in Ba2 and 1.9 10-2 M in Ca2 Ksp(BaSO,) = 1.07 x 10 Kn(CaSO4) = 7.10 x 10-5 What is the ren precipitate? Express youra Ba Submit Pr X Incorrect Provide Feedback rch

Answer 1

Here we need to calculate the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate . Since barium sulfate ( BaSO₄ ) requires less concentration of SO₄^2- ions when compared to calcium sulfate (CaSO₄ ) , Barium sulfate will precipitate first and we need to calculate the concentration of Ba²⁺  cation.

Given , K_sp  of BaSO₄ = 1.07  $\times$ 10^-10

K_sp of CaSO₄  = 7.10 $\times$ 10^-5

  [ Ba²⁺ ] = 1.5 $\times$ 10^-2 M

[ Ca²⁺ ] = 1.9 $\times$ 10^-2 M

BaSO_{4 (s)}  $\rightleftharpoons$ Ba²⁺_(aq) + SO₄^2-_(aq)

K_sp = [ Ba²⁺ ] [ SO₄^2- ]

[ SO₄^2-] = K_sp / [Ba²⁺]

= 1.07 $\times$ 10^-10  / 1.5 $\times$ 10^-2

= 7.133 $\times$ 10^-9  M

CaSO_{4 (s)}$\rightleftharpoons$ Ca²⁺_(aq) + SO₄²⁺_(aq)

Q =  [ Ca²⁺] [ SO₄^2- ]

K_sp = [ Ca²⁺] [ SO₄^2- ]

[ SO₄^2- ] = K_sp / [Ca²⁺]

= 7.10 $\times$ 10^-5 / 1.9 $\times$ 10^-2

= 3.73 $\times$ 10^-3 M

This shows that Barium sulfate requires less concentration of SO₄^2- than calcium sulfate.  

The precipitation just begins when K_sp = Q , that is when K_sp = Q the solution is saturated and started to precipitate

Now we need to find the concentration of  Ba²⁺ ion when Ca²⁺ ion just begins to precipitate ( that is when the concentration of sulfate ion is 3.73 $\times$ 10^-3 M )

K_sp = [ Ba²⁺ ] [ SO₄^2- ]

[Ba²⁺] = K_sp / [SO₄^2-]

= 1.07 $\times$ 10^-10 / 3.73 $\times$ 10^-3

  = 2.868 $\times$ 10^-8 M

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