Can someone help me for this with details Manganese hydroxide (Mn(OH)2) dissolve
ID: 1001729 • Letter: C
Question
Can someone help me for this with details Manganese hydroxide (Mn(OH)2) dissolves sparingly in water as follows; 2+ 20H (aq) (aq) (a) Write the Ks expression for this equation. (1 mark) (b) Given that K 2.1 x 10 13 calculate the concentration of manganese ions and hydroxide ions in a saturated aqueous solution of manganese hydroxide. (2 marks) (c) What is the pH of a saturated solution of the manganese hydroxide? (1 mark) (d) If equal volumes of MnSO4 (0.5 mol L' and NaOH solution (3.05 x 10" mol L were mixed, would you expect to see a precipitate form? Explain your answer. (2 marks) (e) How much Mn with dissolve in a 0.3 mol L solution of KoH? (1 mark)Explanation / Answer
Solution:- the given equation is......
Mn(OH2(s) <---------> Mn2+(aq) + 2OH-(aq)
(a) Here the reactant is solid so will not be involved in equilibrium expression, so Ksp is the product of ion concentrations.
Ksp = [Mn2+] [OH-]2
(b) for calculating the ion concentration for the given Ksp (solubility product) we would make the ice table. No change is considered to the solid and let's say the change for the ion is X.
Mn(OH2(s) <---------> Mn2+(aq) + 2OH-(aq)
I ------ 0 0
C ------- +X +2X
E -------- X 2X
let's plug in the values in Ksp exression..
2.1 x 10-13 = (X)(2X)2 = 4X3
divide both sides by 4
X3 = 5.25 x 10-14
taking cube root on both sides...
X = 3.74 x 10-5
So, the concentration of Mn2+ is 3.74 x 10-5 M and concentration of OH- is 2 x 3.74 x 10-5 M = 7.48 x 10-5 M.
(c) we know that pOH = - log[OH-] = - log(7.48 x 10-5) = 4.13
pH + pOH = 14
So, pH = 14 - 4.13 = 9.87
(d) A precipitate forms if the ionic product is greater than solubility product.
MnSO4(aq) + 2NaOH(aq) ---------> Mn(OH)2(s) + Na2SO4(aq)
let's say we take 1.0L of each then..
moles of MnSO4 that is moles of Mn2+ = 0.5
and moles of NaOH that is OH- = 0.0305
Total volume would be = 1.0 L + 1.0 L = 2.0 L
concentration of Mn2+ in solution = 0.5/2 = 0.25 mol L-
concentration of OH- in solution = 0.0305/2 = 0.01525 mol L-
The equation of these ions could be written as...
Mn2+(aq) + 2OH-(aq) <---------> Mn(OH2(s)
Ionic product = [Mn2+][OH-]2 = 0.25 x (0.01525)2 = 5.81 x 10-5
Ionic product is greater than given Ksp. So, the precipitate of Mn(OH)2 would form.
(e) Mn(OH2 and KOH have one common ion and it is OH-. Due to the presence of common ion, solubility decreases.
let's say the solubility of Mn(OH)2 in presence of KOH is X.
Mn(OH2(s) <---------> Mn2+(aq) + 2OH-(aq)
X X (2X + 0.3)
Ksp = (x) (2X +0.3)2
2.1 x 10-13 = (x) (2X +0.3)2
Ksp value is very low so 2X +0.3 could be taken as 0.3
so, 2.1 x 10-13 = (x) (0.3)2
2.1 x 10-13 = 0.09X
X = 2.1 x 10-13/0.09 = 2.33 x 10-12
So, the solubility of Mn(OH)2 in 0.3 mol L-1 is 2.33 x 10-12 M.
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