1.

Calculate the equilibrium constants of each of the indicated species necessary to reduce an initial 0.2M Zn^(2+) solution to 1.0 xx 10^(-4)Zn^(2+). a. Nh_(3) and Zn(NH_(3))_(4)^(2+) (assume no partial complexation) b. overset(Theta)OH in equilibrium with Zn(OH)_(2)(s). c. overset(Theta)OH and Zn(OH)_(4)^(2-). d. Calculate [overset(Theta)OH] which would be produced by each equilibrium concentration of NH_(3) in part (a). Predict whether Zn(OH)_(2) or Zn(OH)_(4)^(2-) would form in preference to Zn(NH_(3))_(4)^(2+) upon addition of suficient NH_(3) to produce the equilibrium concentration calculated in part(a). e. Explain what would be observeed if concentrated NH_(3) solution were added slowely to 0.2M solution of Zn^(2+). Given. K_(f)Zn(NH_(3))_(4)^(2+) = 5 xx 10^(8). K_(sp)ZN(OH)_(2) = 1.8 xx 10^(-14). K_(f)Zn(OH)_(4)^(2-) = 5 xx 10^(14). K_(b) NH_(4)OH = 1.8 xx 10^(-5).

Answer»

Solution :d. `[Zn^(2+)]` final `=1 xx 10^(-4)M`
`[Zn(NH_(3))_(4)^(2+)]aq = 0.2 - 10^(-4) ~~ 0.2`
`Zn^(2+) + 4NH_(3) HARR Zn(NH_(3))_(4)^(2+)`
`K_(f) = ([Zn(NH_(3))_(4)^(2+)])/([Zn^(2+)][NH_(3)]^(4))`
`5 xx 10^(8) =(0.2)/((10^(-4))[NH_(3)]^(4))`
`(NH_(3))^(4) = 4 xx 10^(-6) = 400 xx 10^(-8)`
`(NH_(3)) = 4.5 xx 10^(-2)M`
b. `Zn(OH)_(2) rarr Zn^(2+) +2 overset(Theta)OH`
`K_(sp) = (1.8 xx 10^(-14))/(10^(-4)) = 1.8 xx 10^(-10)`
`[overset(Theta)OH] = 1.3 xx 10^(-5)M`
C. `Zn^(2+) + 4overset(Theta)OH rarr Zn(OH)_(4)^(2-)`
`K_(f) - ([Zn(OH)_(4)^(2-)])/([Zn^(2+)][overset(Theta)OH]^(4))`
`5 xx 10^(4) = (0.2)/((10^(-4))[overset(Theta)OH]^(4))`
`rArr [overset(Theta)OH]^(4) = 4 xx 10^(-12), [overset(Theta(O)H] = 1.4 xx 10^(-3)`
d. The concentration of `overset(Theta)OH` in equilibrium with the `NH_(3)` concentration of part (a) is given by,
`NH_(3) + H_(2)O hArr overset(o+)NH_(4) + overset(Theta)OH`,
(LET `x = [overset(o+)NH_(4)] = [overset(Theta)OH]`
`K_(b) = ([overset(o+)NH_(4)][overset(Theta)OH])/([NH_(3)])`
`1.8 xx 10^(-5) = (x^(2))/(4.5 xx 10^(-2))`
`rArr x^(2) = 8.1 xx 10^(-7) rArr x = [overset(Theta)OH] = 9.0 xx 10^(-4)M`.
`[overset(Theta)OH] (9.0 xx 10^(-4)M)` from the `4.5 xx 10^(-2)M NH_(3)` is sufficient precipitate `Zn(OH)_(2)` [part (b)] but not sufficient to from `Zn(OH_(4)^(2-))` [part (c)]
e. On addition of `Nh_(3)` solution to `Zn^(2+)` solution, a precipitate of `Zn(OH)_(2)` will form, which will later dissolve to yield a clear solution containing `[Zn(NH_(3))_(4)]^(2+)`


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