Manganese hydroxide (Mn(OH)_2) dissolves sparingly in water as follows; Mn(OH)_2

Question

Manganese hydroxide (Mn(OH)_2) dissolves sparingly in water as follows; Mn(OH)_2(s) Mn^2+_(aq) + 2OH^-_(aq) Write the K_s expression for this equation. Given that K_s = 2.1 times 10^-13, calculate the concentration of manganese ions and hydroxide ions in a saturated aqueous solution of manganese hydroxide. What is the pH of a saturated solution of the manganese hydroxide? If equal volumes of MnSO_4 (0.5 mol L^-1) and NaOH solution (3.05 times 10^-2 mol L^-1) were mixed, would you expect to see a precipitate form? Explain your answer. How much Mn(OH)_2 with dissolve in a 0.3 mol L^-1 solution of KOH?

Explanation / Answer

Mn (OH)2 dissolves sparingly in water as follows:

Mn (OH)2 (s) < == > Mn2+    (aq) + 2OH- (aq)

the Ks expression for this reaction is as follows:

Ks = [Mn2+] [OH-]^2

Assume the concenrtaion of Mn 2+ = x and OH- = 2X

Ksp = 2.1x 10^-13 = x ( 2x)^2 = 4x^3

2.1x 10^-13= 4x^3

X^3 = 5.25*10^-14

x = 3.74 x 10^-5 mol /L =[Mn2+]
[OH-] = 2 x 3.74 x 10^-5

=7.48 x 10^-5 M
pOH = - log 3.4 x 10^-5 =4.5
pH = 14 - 4.5 = 9.5

2NaOH (aq)+MgSO4 (aq) ----> Mg(OH)2 (s) + Na2SO4 (aq)

2Na+(aq) + 2OH-(aq) + Mg+2(aq) + (SO4)-2(aq) --> Mg(OH)2 (s) + 2Na+(aq) +SO4)-2 (aq)

Because 2NaOH ; MgSO4 (aq) and Na2SO4 (aq) are ionic compound.

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

A precipitate will form when two solutions are mixed and the the ion product is greater than the solubility product which is expressed as follows:

[M+][A-] > Ksp

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