Question

please help me step by step!!

3.) The solubility of Ag2SO4 (a solid) in water is 1.22x10^-2 g/L. Calculate the solubiity product constant (Ksp) for this substance.

4.) in the tritration of a week acid with strong base, the pH will change very gradually, then change very rapidly near the equivalence point. Draw a rough sketch of the graph and explain.

Answer 1

3)

Molar mass of Ag2SO4,

MM = 2*MM(Ag) + 1*MM(S) + 4*MM(O)

= 2*107.9 + 1*32.07 + 4*16.0

= 311.87 g/mol

Molar mass of Ag2SO4= 311.87 g/mol

s = 1.22*10^-2 g/L

To covert it to mol/L, divide it by molar mass

s = 1.22*10^-2 g/L / 311.87 g/mol

s = 3.912*10^-5 mol/L

At equilibrium:

Ag2SO4 <----> 2 Ag+ + SO42-

   2s s

Ksp = [Ag+]^2[SO42-]

Ksp = (2s)^2*(s)

Ksp = 4(s)^3

Ksp = 4(3.912*10^-5)^3

Ksp = 2.395*10^-13

Answer: 2.40*10^-13

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