Question

Find pH.
A) 0.10M Zn(OH)4^-2

F) Zn(OH)2 in 0.010M ZnCl2


Q) 0.010 M Zn(NH3)4^+2 hint: fined [NH3] then kb for OH^-


Im having trouble with these can someone help me.

Answer 1

For a, I would write
..........Zn(OH)4^- ==> Zn^2+ + 4OH^-
I.........0.10.............0......0
C........-0.10...........0.1...+0.4
E..........0.............+0.1...0.4
Convert OH^- to pOH, then to pH.

For b.
Let x = solubility Zn(OH)2
.........Zn(OH)2 ===> Zn^2+ + 2OH^-
E..........x...........x......2x..
Ksp = (Zn^2+)(OH^-^2
For Zn^2+ substitute x + 0.01M (x from Zn(OH)2 and 0.01 from ZnCl2.)
Solve for x, convert to 2x = OH^-, then convert to pH.

For c, write the equations.
............Zn^2+ + 4NH3 ==> Zn(NH3)4^2+
Kf = [Zn(NH3)4^2+]/(Zn^2+)(NH3)^4
NH3 + HOH ==> NH4^+ + OH^-
Follow the hint: You know Kf and the concn of the complex ion, solve for NH3 (4x) and Zn(x). Then plug that (NH3) into the NH3 equation and solve for OH^- and convert to pH.