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Need help all of them What is the solubility product K_apt of Mg(OH)_2 if its so

ID: 1071104 • Letter: N

Question

Need help all of them
What is the solubility product K_apt of Mg(OH)_2 if its solubility in water is 1.6 times 10^-4 mol L^-1? a) 1.6 times 10^-11 B) 2.6 times 10^-4 C) 3.2 times 106-4 D) 4.1 times 10^-11 A saturated solution of MgF_2 contains 1.6 times 10^-3 mol of MgF_2 per liter at a certain temperature. What is the K_aps of MgF_2 at this temperature? A) 2.7 times 10^-4 B) 1.6 times 10^-8 C) 3.1 times 10^-9 D) 6.2 times 10^-8 What will be the result if 100 mL of 0.06 M Mg(NO)_2 is added to 50 mL of 0.06 M Na_2C_2O_4? Assume the reaction is taking place at 25 degree C. A) No precipitate will form. B) A precipitate will form and an excess of Mg^2+ ions will remain in solution. C) A precipitate will form and an excess of C_2O_4^2- ion swill remain in solution. D) A precipitate will from but neither ion is present in excess. The addition of solid Na_2SO_4 to an aqueous solution in equilibrium with solid BaSO_4 will cause A) no change in [Ba^2+] in solution. B) more BaSO_4 to dissolve. C) precipitation of more BaSO_4. D) an increase in the K_ap of BaSO_4 In a 0.050 M solution of a weak monoprotic acid, [H^+] = 1.8 times 10^-3. What is its K_a? A) 3.6 times 10^-2 B) 9.0 times 10^-5 C) 6.7 times 10^-5 D) 1.6 times 10^-7 M solution of a weak acid is found to be 1.3% ionized. What is its K_a? 1.3 times 10^-2 B) 2.0 times 10^-3 C) 1.1 times 10^-3 D) 2.6 times 10^-5

Explanation / Answer

Q1

Solubility Product, Ksp for

Mg(OH)2 given

S = 1.6*10^-4 mol /L

so

Mg(OH)2(s) = Mg+2(aq) + 2OH-(aq)

note that for 1 mol of Mg(OH)2 we have:

1 mol of Mg+2 ions

2 mol of OH- ions

so

[Mg+2] = S

[OH-] = 2S

so

Ksp = [Mg+2][OH-]^2

Ksp = (s)(2S)^2

4*S^3 = Ksp

4(1.6*10^-4)^3 = Ksp

Ksp= 1.638*10^-11

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