Chemistry, 28.02.2020 00:14 milo3685

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)

Answers: 2

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Answers

Answer from: JanaMiqdad1003

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.

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.

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.

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.

Answer from: Quest

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2

Answer from: Quest

The metal is named first

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For which of the following mixtures will Ag2SO4(s) precipitate?
a. 150.0 mL of 0.10 M Na2SO4(a...