Answer:
c. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)
d. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)
Explanation:
Dissociation of Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
Solubility Product ; =
It is known that the solubility product of Ag₂SO₄ = 1.2 × 10⁻⁵
If ionic concentration < ; precipitation will not occur
If ionic concentration > ; precipitation will occur
Now, let's pick the option one after the other to determine if precipitation occurs or not.
a).
150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)
moles of Ag =
= 0.001 moles
moles of =
= 0.015 moles
New total volume = 150.0 mL + 5.0 mL = 155 mL
= 0.155 L
Molar Concentration of Ag:
=
=
Molar Concentration of
=
=
Ionic Concentration =
= ×
=
∴ < ; therefore precipitation will not occur.
b)
150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)
moles of Ag =
= 0.0015 mole
moles of =
= 0.015 moles
Molar Concentration of Ag:
=
=
Molar Concentration of
=
=
Ionic Concentration =
=
=
=
∴ < ; therefore precipitation will not occur.
c)
150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)
moles of Ag =
= 0.002 moles
moles of =
= 0.015 moles
Molar Concentration of Ag:
=
=
Molar Concentration of
=
=
Ionic Concentration =
=
=
∴ > ; therefore precipitation will occur.
d)
150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)
moles of Ag =
= 0.0025 moles
moles of =
= 0.015 moles
Molar Concentration of Ag:
=
=
Molar Concentration of
=
=
Ionic Concentration =
=
=
∴ > ; therefore precipitation will occur.