Question

K2SO4(aq) + AgNO3(aq) → Ag2SO4(s) + KNO3 (aq)

a Balance the reaction.

b. Given that you have a 1.75 L solution of 0.170 M K2SO4 and a 2.00 L solution of 0.110 M AgNO3, calculate the amount in grams of precipitate formed. Show work

c.Calculate the concentration of SO2− ions left after the reaction has been completed

precipitate formed. show work

Answer 1

(a) Balancing the equation

K₂SO₄(aq) + AgNO₃(aq) → Ag₂SO₄(s) + KNO₃ (aq)

You can see that the starting material has 2 K atoms so the product needs to have 2 K so 1 KNO₃ become 2 KNO₃ now since NO₃ is 2 equivalent we make AgNO₃ in the starting as 2 AgNO₃. So now there are 2 Ag to start with but in the product also we have 2 Ag as Ag₂SO₄ so the equation is completelt balanced

Balanced equation

K₂SO₄(aq) + 2AgNO₃(aq) → Ag₂SO₄(s) + 2KNO₃ (aq)

(b) 1.75 L solution of 0.170 M K₂SO₄ and a 2.00 L solution of 0.110 M AgNO₃

1.75 L solution of 0.170 M K₂SO₄

means 1.75 x 0.170 M = 0.2975 moles of K₂SO₄  

2.00 L solution of 0.110 M AgNO₃

means 2.0 x 0.110 M = 0.22 moles of  AgNO₃

As per the balance equation

K₂SO₄(aq) + 2AgNO₃(aq) → Ag₂SO₄(s) + 2KNO₃ (aq)

2 moles of AgNO₃ gives 1 mole of  Ag₂SO₄(s) which is the precipitate

since we have 0.22 moles of  AgNO₃ we will have 0.11 mole of Ag₂SO₄(s) precipitate

MW of Ag₂SO₄ is 311.79 g·mol⁻¹

so 0.11 mole of Ag₂SO₄ will be 0.11 moles x 311.79 g·mol⁻¹ = 34.29 g

so  34.29 g of precipitate is formed

(c) Calculate the concentration of SO2− ions left after the reaction has been completed

0.11 mole of Ag₂SO₄(s) precipitate

1 mole of Ag₂SO₄ has 1 mole of SO₄ (sulphate)

We started the reaction with 0.2975 moles of K₂SO₄ this has 0.2975 moles of SO₄

since 0.11 moles of SO₄ precipitate as  Ag₂SO₄

0.2975 - 0.11 moles of SO₄ remain in solution

0.1875 moles of SO₄ remain in solution

In the reaction total volume is 1.75 L K₂SO₄ solution and a 2.00 L AgNO₃ solution

That is 3.75 L total solution

Concentration of SO₄ will be moles/L or  0.1875 moles in 3.75 L

That is 0.05 moles/litre or 0.05 M is the concentration of sulphate