CHEMISTRY EQUILIBRIUM PROBLEM! 6M NH_3 is added to a Ni(OH)_2 precipitate so tha
ID: 972817 • Letter: C
Question
CHEMISTRY EQUILIBRIUM PROBLEM!
6M NH_3 is added to a Ni(OH)_2 precipitate so that some of the precipitate dissolves according to the equation: Ni(OH)_2 + 6 NH_3 reverseequilibrium Ni(NH_3)_6^2+(aq) + 2 OH^-(aq) What does the NH_3 concentration need to be at equilibrium in order to dissolve enough nickel(II) hydroxide such that there is 0.10M Ni(NH_3)_6^2+ in solution? K_sp(Ni(OH)_2) = 5.5 times 10^-16; K_f(Ni(NH_3)_6^2+) = 5.6 times 10^8. K = (5.5 times 10^-16)(5.6 times 10^8) = 3.08 times 10^-7 3.08 times 10^-7 = [Ni(NH_3)_6^2+][OH^-]^2/[NH_3]^6 Ni(OH)_2 (s) + 6HN_3(aq) reverseEquilibrium Ni(NH_3)_6^2+(aq) + 2OH^-(aq)Explanation / Answer
Data.
[NH3] = 6 M add to Ni(OH)2 (s)
[NH3] = ?? to precipitate Ni in a 0.10 M Ni(NH3)6+2 dissolution
Kps = 5.5x10-16
Kf = 5.6x108
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Ammonia reacts with water forming OH- ions, which in turn reacts with Ni+2 to form Ni(OH)2. The interest equilibriums are:
NH3 (aq) + H2O (l) <==> NH4+ (ac) + OH- (ac)
Ni2+ (ac) + 2OH- (ac) <==> Ni(OH)2 (s)
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First, we find the [OH-] above which it begins to precipitate Ni(OH)2
Kps = [Ni2+][OH-]2 = 5.5x10-16
since, Ni(NH3)6+2 is a strong electrolite, [Ni2+] = 0.10 M
[OH-]2 = 5.5x10-16 /0.10
[OH-] = 7.41x10-8 M
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Now we calculate [NH3] which will provide 7.41x10-8 M of OH- ions. Let x be the [NH3]o in moles/L:
NH3 (ac) + H2O (l) <==> NH4+ (ac) + OH- (ac)
(x - 7.41x10-8) ----------- (+7.41x10-8) . (+7.41x10-8)
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Then,
Kb = [NH4+][OH-]/[NH3]
1.8x10-5 = (+7.41x10-8)2/(x - 7.41x10-8)
x = 4.12x10-3 M
Therefore, the concentration of ammonia needed, [NH3] should be enough than "x" to start the precipitation of Ni(OH)2.
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From chemical reaction we know:
Ni2+ (ac) + 6 NH3 (ac) <==> Ni(NH3)62+ (aq)
Kf = [Ni(NH3)6+2]/[Ni2+][NH3]6
5.6x108 = 0.10/[Ni2+](4.12x10-3)6
[Ni2+] = 4.33x10-8 M
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The concentration of nickel (II) ions is 4.33x10-8 M. This approximation is justified, becuase [Ni2+] in the equilibrium is too small compared to 0.10 M.