Question

1. How many grams of CaCO₃ will dissolve in 2.80 x10² ml of 0.040 M of Ca(NO₃)₂? The K_sp for the CaCO₃ is 8.7 x 10^-9. Answer must be in scientific notation.

2. Calculate the molar solubility of AgCl in a 1.00 L solution containing 20.0 g of dissolved CaCl₂? Answer must be in scientific notation.

Answer 1

1) CaC03 ---> Ca+2 + C03-

now

using ICE table

initial conc of Ca+2 , C032- are 0.04 , 0

change in conc of Ca+2 , C032- are + x , +x

equilibrium conc of Ca+2 , C032- are 0.04 + x , x

now

Ksp = [Ca+2] [C032-]

so

Ksp = [0.04 +x ] [x]

[0.04+x] [x] = 8.7 x 10-9

solving we get

x = 2.175 x 10-7 M

now

moles of CaC03 = molarity x volume (L)

moles of CaC03 = 2.175 x 10-7 x 0.28

moles of CaC03 = 6.09 x 10-8

now

mass = moles x molar mass

so

mass of caC03 = 6.09 x 10-8 x 100

so

6.09 x 10-6 grams of CaC03 will dissolve

2)

we know that

moles = mass / molar mass

so

moles of CaCl2 = 20 / 111

moles of CaCl2 = 0.18018

now

molarity = moles / volume (L)

so

molarity of CaCl2 = 0.18018 / 1 = 0.18018 M

now

CaCl2 ---> Ca+2 + 2Cl-

so

{Cl-] = 2 [CaCl2]

[Cl-] = 2 x 0.18018

[Cl-] = 0.36036 M

we know that

Ksp of AgCl = 1.8 x 10-10

now

AgCl ---> Ag+ + Cl-

Ksp = [Ag+] [Cl-]

so

1.8 x 10-10 = [Ag+] [ 0.36036]

[Ag+] = 4.995 x 10-10

so

the molar solubility of AgCl is 4.995 x 10-10 M