Question

3. What is the pH as precipitation of nickel hydroxide just begins from a 0.010 M nickel sulfate solution?

4. Will a precipitate form when 1.0 mL of 1.0 M potassium sulfate solution, 10.0 mL of 0.0030 M calcium chloride solution, and 100.0 mL of deionized water are mixed? Show all supporting calculations for your decision, including the identity of the possible precipitate(s).

Answers are:

3) 7.36

4) no ppt of CaSO₄, Q = 2.4 x 10^-6

^{I need soutions.}

Answer 1

0.010 M Nickel sulphate will dissociate in solution to give 0.010 M Ni2+ and 0.010 M of SO₄^2-.

Nickel hydroxide will start to precipitate from the solution when the ionic product of Ni²⁺ and OH^- will exceed the solubility product (Ksp)

Ni 2+   + 2OH- ------------------------> Ni (OH)₂

ion product = [Ni2+][OH-]²

Ksp of Nickel hydroxide = 5.48 * 10^-16 (refer your text book)

[Ni2+][OH-]²= 5.48 * 10^-16

[OH^-] ²= 5.48 * 10^-16 / 0.010 = 5.48 * 10^-14

[OH]= 2.34 * 10^-7

pOH= - log [OH-]

      = - log [ 2.34 * 10^-7]

      = 6.63

pH = 14-pOH

     = 14 -6.63

     = 7.37 (Ans)

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Total volume = 1 + 10 + 100 = 111 mL

Molarity of K₂SO₄ after mixing

M1V1= M2V2

M1 = M2V2/ v1 = 1 * 1/111 = 0.009 M

Molarity of CaCl2 after mixing

M1V1 = M2V2

M1 = M2V2/ V1

     = 10 * 0.003/ 111 = 2.7 * 10^-4 M

K₂SO₄ -------------> 2K ⁺   +             SO₄^2-

0.009 M               2 * 0.009M              0.009 M

concentration of SO₄^2- = 0.009 M

CaCl2 ------------> Ca2+    +    2Cl-

2.7 * 10^-4 M           2.7 * 10^-4 M                  2 * 2.7 * 10^-4 M

concentration of Ca²⁺ = 2.7* 10^-4 M

The precipitate of CaSO4 will form if the ion product exceeds solubility product

ion product of CaSO4 (Q)= [Ca²⁺][SO₄^2-]

                                = 2.7* 10^-4 M * 0.009 M

                                =0.024 * 10^-4

                                           = 2.4 * 10^-6

Solubility product of CaSO4 = 4.93 x 10^-5 (refer your text book) which is more than the ion product.

No precipitate of caSO4 will form

No