Consider an amphoteric hydroxide, M(OH)2(s). where M is a generic metal. Estimat

Question

Consider an amphoteric hydroxide, M(OH)2(s). where M is a generic metal. Estimate the solubility of M(OH)2 in a solution buffered at pH = 7.0, 10.0, and 14.0. Copper(l) ions in aqueous solution react with NH3(ag) according to Calculate the solubility (in g- L-1) of CuBr(s) (Kgp = 6.3 times 10-9) in 0.88 M NH3(aq). The formation constant of [M(CN)2] is 5.30 times 1018, where M is a generic metal. A 0.160-mole quantity of M(NC>3) is added to a liter of 0.710 M NaCN solution. What is the concentration of M + ions at equilibrium?

Explanation / Answer

Ksp =2 x l0^-16th = (M++)(OH)^2
at pH=7
pH + pOH = 14
pOH= 7
(OH-) = antilog(-pOH)
(OH-) = antilog (-7)
= l x l0^-7
2 x l0^-16 = (M+2)(l x l0^-7)^2
(M+2) = 2 x l0^-16/1 x l0^-14
Solubility of M(OH)2 at pH 7 = 2 x l0^-2


If pH = l0
pOH = 4
(OH-) = antilog(-pOH)
antilog (-4) = l x l0^-4
2 x l0^-16 = (M2+)(l x l0^-4)^2
(M2+) = 2 x l0^-16/1 x l0^-8
Solubility of M(OH)2 at pH 10 = 2 x l0^-8

at pH=14
pOH= 0
(OH-) = antilog(-pOH)
antilog (-0) = 1
2 x l0^-16 = (Mg++)1^2
(Mg++) = 2 x l0^-16/1
Solubility of M(OH)2 at pH 14 = 2 x l0^-16

CuBr ==> Cu^+ + Br^-
Cu^+ + 2NH3 ==> Cu(NH3)2^+
----------------------------
add equn 1 to eqn 2.
   CuBr(s) + 2NH3 ==>Cu(NH3)2^+ + Br^-
I          0.88M     0          0
E           0.88-2x. x       

Ksp= [Cu(NH3)2]* [Br^-] = s^2
6.3*10^-9 = S^2
S= ?6.3*10^-9
Solubility = 7.9*10^-5M

Solubility in g/l = 3.5*10^-7 g/l

M^2+(aq) + 2CN^- ? [M(CN)2]^1-
Kform = [[M(CN)2]^1-]/[ M^2+][CN^-]^2 = 5.30

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