Question

For which of the following mixtures will Ag2SO4(s) precipitate?

Answer 1

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 ;   $K_{sp$ = $[Ag^+]^2[SO_4^{2-}]$

It is known that the solubility product   $K_{sp$ of Ag₂SO₄ = 1.2 × 10⁻⁵

If ionic concentration $K_I$ < $K_{sp}$  ; precipitation will not occur

If  ionic concentration $K_I$ > $K_{sp}$ ; 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 = $\frac{5*1}{1000}*0.20M$

= 0.001 moles

moles of $SO_4^{2-$ = $\frac{150*1}{1000} *0.10M$

= 0.015 moles

New total volume = 150.0 mL + 5.0 mL = 155 mL

= 0.155 L

Molar Concentration of Ag:

$[Ag^+]^2$ = $\frac{0.001}{0.155}$

= $6.45*10^{-3}$

Molar Concentration of   $SO_4^{2-$

$[SO_4^{2-}]$ = $\frac{0.015}{0.155}$

= $96.77*10^{-3}$

Ionic Concentration $K_I$  = $[Ag^+]^2$  $[SO_4^{2-}]$

= $(6.45*10^{-3})^2$ × $96.77*10^{-3}$

= $4.02*10^{-6}$

∴  $K_I$  <  $K_{sp$ ; 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 = $\frac{5*1}{1000}*0.30M$

= 0.0015 mole

moles of $SO_4^{2-$ = $\frac{150*1}{1000} *0.10M$

= 0.015 moles

Molar Concentration of Ag:

$[Ag^+]^2$ = $\frac{0.0015}{0.155}$

= $9.677*10^{-3}$

Molar Concentration of   $SO_4^{2-$

$[SO_4^{2-}]$ = $\frac{0.015}{0.155}$

= $96.77*10^{-3}$

Ionic Concentration $K_I$  = $[Ag^+]^2$  $[SO_4^{2-}]$

= $(9.364*10^{-5})^2*(96.77*10^{-3})$

= $9.057672*10^{-6}$

= $9.06*10^{-6}$

∴  $K_I$  <  $K_{sp$ ; 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 = $\frac{5*1}{1000}*0.40M$

= 0.002 moles

moles of $SO_4^{2-$ = $\frac{150*1}{1000} *0.10M$

= 0.015 moles

Molar Concentration of Ag:

$[Ag^+]^2$ = $\frac{0.002}{0.155}$

= $12.9*10^{-3}$

Molar Concentration of   $SO_4^{2-$

$[SO_4^{2-}]$ = $\frac{0.015}{0.155}$

= $96.77*10^{-3}$

Ionic Concentration $K_I$  = $[Ag^+]^2$  $[SO_4^{2-}]$

= $(12.9*10^{-3})^2(96.77*10^{-3})$

= $1.61*10^{-5}$

∴  $K_I$  >  $K_{sp$ ; 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 = $\frac{5*1}{1000}*0.50M$

= 0.0025 moles

moles of $SO_4^{2-$ = $\frac{150*1}{1000} *0.10M$

= 0.015 moles

Molar Concentration of Ag:

$[Ag^+]^2$ = $\frac{0.0025}{0.155}$

= $16.1*10^{-3}$

Molar Concentration of   $SO_4^{2-$

$[SO_4^{2-}]$ = $\frac{0.015}{0.155}$

= $96.77*10^{-3}$

Ionic Concentration $K_I$  = $[Ag^+]^2$  $[SO_4^{2-}]$

= $(16.1*10^{-3})^2(96.77*10^{-3})$

= $2.508*10^{-5}$

∴  $K_I$  >  $K_{sp$ ; therefore precipitation will occur.