For which of the following mixtures will Ag2SO4(s) precipitate?
a. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)
b. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)
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)

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.