Question

I have two questions that I do not know how to answer. Thanks for your help!

1. Calculate the [OH^-], [H^+], pH, and pOH for a 0.50M solution of CN^-

2. Calculate the molar solubility of PbCl2 in 0.40M KCl

Answer 1

The pH et al are determined by the hydrolysis of the CN^-.
........CN^- + HOH ==> HCN + OH^-
I......0.5..............0.....0
C.......-x..............x.....x
E.....0.5-x.............x.....x

Kb for CN = (Kw/Ka for HCN) = (x)(x)/(0.5-x)

You know or can look up Kw for H2O and Ka for HCN. Then solve for x and that gives you OH^- directly. I would convert that to pOH, then use
pH + pOH = pKw = 14 and solvle for pH.
Finally, since pH = =log(H^+) and you now know pH, you can solve for (H^+).


For #2 there are two equilibria. The KCL is 100% ionized so
.........KCl ==> K^ + Cl^-
I........0.4......0....0
C........-0.4....0.4...0.4
E.........0......0.4....0.4

PbCl2 is a slightly soluble material and will have a Ksp. You didn't provide that so you must find it in the problem or look it up in tables.
..........PbCl2 --> Pb^2+ + 2Cl^-
I..........solid....0........0
C,.........solid....x........2x
E..........solid....x........2x

Ksp = (Pb^2+)(Cl^-)^2
You know or can look up Ksp.
(Pb^2+) = x
(Cl^-) = 2x from the PbCl2 and 0.4 from the KCl so (Cl^-) = 0.4+2x

Substitute those values into the Ksp expression and solve for (Pb^+) = x = solubility in mols/L.
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