A solution contains o.036 M Cu^2+ and o.044 M Fe^2+. A solution containing sulfi

Question

A solution contains o.036 M Cu^2+ and o.044 M Fe^2+. A solution containing sulfide ions is added to selectively precipitate one of the metal ions from solution. At what concentration of sulfide ion will a precipitate begin to form? What is the identity of the precipitate? K_sp(CuS) = 1.3 times 10^-36, K_sp(FeS) = 6.3 times 10^-18 What is the identity of the precipitate? At what concentration of sulfide ion ill a precipitate begin to form? What is the concentration of the anion when the second action begins to precipitate? The molar solubility of iron (Ill) hydroxide in pure water at 25 degree C is 9.94 times 10^-10 M. How would the substance's molar solubility change if it is placed in an aqueous solution of pH 10.0 at 25 degree C? Support your answer with appropriate calculations.

Explanation / Answer

(a) CuS precipitate is fromed first and then FeS is formed.

The identity is a black colour of CuS is formed first.

(b)

CuS (s) = Cu2+ (aq.) + S2- (aq.)

Ksp = [Cu2+][S2-]

1.3 * 10-36 = 0.036 * [S2-]

[S2-] = 3.61 * 10-35 M

Concentration of anion when first precipitate begins to form = 3.61 * 10-35 M

(c)

FeS (s) = Fe2+ (aq.) + S2- (aq.)

Ksp = [Fe2+][S2-]

6.3 * 10-18 = 0.044 * [S2-]

[S2-] = 1.43 * 10-16 M

Concentration of anion when second action begins to precipitate = 1.43 * 10-16 M

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