Question

please show how you get an answer please

1. Determine the solubility in g/L of each of the following salts: (a) AgCl, K, = 1.6 x 10" (b) AgBr, K = 5.0 x 10" (c) Ag,Cro,, K = 9,0 x 10-12

Answer 1

a)

At equilibrium:

AgCl <----> Ag+ + Cl-

   s s

Ksp = [Ag+][Cl-]

1.6*10^-10=(s)*(s)

1.6*10^-10= 1(s)^2

s = 1.265*10^-5 M

Molar mass of AgCl,

MM = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

Molar mass of AgCl= 143.35 g/mol

s = 1.265*10^-5 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.265*10^-5 mol/L * 143.35000000000002 g/mol

s = 1.813*10^-3 g/L

Answer: 1.8*10^-3 g/L

b)

At equilibrium:

AgBr <----> Ag+ + Br-

   s s

Ksp = [Ag+][Br-]

5*10^-13=(s)*(s)

5*10^-13= 1(s)^2

s = 7.071*10^-7 M

Molar mass of AgBr,

MM = 1*MM(Ag) + 1*MM(Br)

= 1*107.9 + 1*79.9

= 187.8 g/mol

Molar mass of AgBr= 187.8 g/mol

s = 7.071*10^-7 mol/L

To covert it to g/L, multiply it by molar mass

s = 7.071*10^-7 mol/L * 187.8 g/mol

s = 1.328*10^-4 g/L

Answer: 1.3*10^-4 g/L

c)

At equilibrium:

Ag2CrO4 <----> 2 Ag+ + CrO42-

   2s s

Ksp = [Ag+]^2[CrO42-]

9*10^-12=(2s)^2*(s)

9*10^-12= 4(s)^3

s = 1.31*10^-4 M

Molar mass of Ag2CrO4,

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

= 2*107.9 + 1*52.0 + 4*16.0

= 331.8 g/mol

Molar mass of Ag2CrO4= 331.8 g/mol

s = 1.31*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.31*10^-4 mol/L * 331.8 g/mol

s = 4.348*10^-2 g/L

Answer: 4.3*10^-2 g/L