Respuesta :
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 ; [tex]K_{sp[/tex] = [tex][Ag^+]^2[SO_4^{2-}][/tex]
It is known that the solubility product [tex]K_{sp[/tex] of Ag₂SO₄ = 1.2 × 10⁻⁵
If ionic concentration [tex]K_I[/tex] < [tex]K_{sp}[/tex] ; precipitation will not occur
If ionic concentration [tex]K_I[/tex] > [tex]K_{sp}[/tex] ; 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 = [tex]\frac{5*1}{1000}*0.20M[/tex]
= 0.001 moles
moles of [tex]SO_4^{2-[/tex] = [tex]\frac{150*1}{1000} *0.10M[/tex]
= 0.015 moles
New total volume = 150.0 mL + 5.0 mL = 155 mL
= 0.155 L
Molar Concentration of Ag:
[tex][Ag^+]^2[/tex] = [tex]\frac{0.001}{0.155}[/tex]
= [tex]6.45*10^{-3}[/tex]
Molar Concentration of [tex]SO_4^{2-[/tex]
[tex][SO_4^{2-}][/tex] = [tex]\frac{0.015}{0.155}[/tex]
= [tex]96.77*10^{-3}[/tex]
Ionic Concentration [tex]K_I[/tex] = [tex][Ag^+]^2[/tex] [tex][SO_4^{2-}][/tex]
= [tex](6.45*10^{-3})^2[/tex] × [tex]96.77*10^{-3}[/tex]
= [tex]4.02*10^{-6}[/tex]
∴ [tex]K_I[/tex] < [tex]K_{sp[/tex] ; 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 = [tex]\frac{5*1}{1000}*0.30M[/tex]
= 0.0015 mole
moles of [tex]SO_4^{2-[/tex] = [tex]\frac{150*1}{1000} *0.10M[/tex]
= 0.015 moles
Molar Concentration of Ag:
[tex][Ag^+]^2[/tex] = [tex]\frac{0.0015}{0.155}[/tex]
= [tex]9.677*10^{-3}[/tex]
Molar Concentration of [tex]SO_4^{2-[/tex]
[tex][SO_4^{2-}][/tex] = [tex]\frac{0.015}{0.155}[/tex]
= [tex]96.77*10^{-3}[/tex]
Ionic Concentration [tex]K_I[/tex] = [tex][Ag^+]^2[/tex] [tex][SO_4^{2-}][/tex]
= [tex](9.364*10^{-5})^2*(96.77*10^{-3})[/tex]
= [tex]9.057672*10^{-6}[/tex]
= [tex]9.06*10^{-6}[/tex]
∴ [tex]K_I[/tex] < [tex]K_{sp[/tex] ; 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 = [tex]\frac{5*1}{1000}*0.40M[/tex]
= 0.002 moles
moles of [tex]SO_4^{2-[/tex] = [tex]\frac{150*1}{1000} *0.10M[/tex]
= 0.015 moles
Molar Concentration of Ag:
[tex][Ag^+]^2[/tex] = [tex]\frac{0.002}{0.155}[/tex]
= [tex]12.9*10^{-3}[/tex]
Molar Concentration of [tex]SO_4^{2-[/tex]
[tex][SO_4^{2-}][/tex] = [tex]\frac{0.015}{0.155}[/tex]
= [tex]96.77*10^{-3}[/tex]
Ionic Concentration [tex]K_I[/tex] = [tex][Ag^+]^2[/tex] [tex][SO_4^{2-}][/tex]
= [tex](12.9*10^{-3})^2(96.77*10^{-3})[/tex]
= [tex]1.61*10^{-5}[/tex]
∴ [tex]K_I[/tex] > [tex]K_{sp[/tex] ; 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 = [tex]\frac{5*1}{1000}*0.50M[/tex]
= 0.0025 moles
moles of [tex]SO_4^{2-[/tex] = [tex]\frac{150*1}{1000} *0.10M[/tex]
= 0.015 moles
Molar Concentration of Ag:
[tex][Ag^+]^2[/tex] = [tex]\frac{0.0025}{0.155}[/tex]
= [tex]16.1*10^{-3}[/tex]
Molar Concentration of [tex]SO_4^{2-[/tex]
[tex][SO_4^{2-}][/tex] = [tex]\frac{0.015}{0.155}[/tex]
= [tex]96.77*10^{-3}[/tex]
Ionic Concentration [tex]K_I[/tex] = [tex][Ag^+]^2[/tex] [tex][SO_4^{2-}][/tex]
= [tex](16.1*10^{-3})^2(96.77*10^{-3})[/tex]
= [tex]2.508*10^{-5}[/tex]
∴ [tex]K_I[/tex] > [tex]K_{sp[/tex] ; therefore precipitation will occur.
The mixture of Ag2SO4(s) that will precipitate:
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)
Precipitation
This is known that Ag2SO4(s) dissociateds when:
[tex]Ag_{2} SO_{4}[/tex](s) ⇄ [tex]2Ag^{+}[/tex] + [tex]SO_{4}^{2-}[/tex]
The solubility [tex]K_{sp}[/tex] of [tex]Ag_{2} SO_{4}[/tex] [tex]= 1.2[/tex] × [tex]10^-5[/tex]
Also,
The precipitation will take place only when:
[tex]K_{1}[/tex] > [tex]K_{sp}[/tex]
so,
by measuring [tex]K_{sp}[/tex] of each of the given mixtures. we find that mixtures having [tex]150.0 ml[/tex] of [tex]0.10 M[/tex] [tex]Na2SO4(aq)[/tex] and [tex]5.0 mL[/tex] of [tex]0.40 M[/tex][tex]AgNO3(aq)[/tex]
and
mixture with [tex]150.0 ml[/tex] of [tex]0.10 M[/tex] [tex]Na2SO4(aq)[/tex] and [tex]5.0 ml[/tex] of [tex]0.50 M[/tex] of [tex]AgNO3(aq)[/tex] would precipitate.
Thus, options c and d are the correct answers.
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