Question

Use the following equation: Ag2SO4 + 2KCl → 2AgCl + K2SO4

How many mL of 1.4 M Ag2SO4 are needed to titrate out 75.0 mL of 2.5 M KCl? How many L of 0.250 M KCl are needed to titrate 72.8 g of AgCl? If 94.1 mL of 0.762 M KCl is required to titrate 61.8 mL of Ag2SO4 what is the molarity of the Ag2SO4? If 62.8 mL of 0.236 M Ag2SO4 is titrated with 23.7 mL of KCl to reach the end point, what is the molarity of the KCl?

Answer 1

A.

Ag₂SO₄ + 2KCl → 2AgCl + K₂SO₄

Here 1 mole of Ag₂SO₄ reacts with 2 moles of KCl.

2.5 M KCl

In 1000 mL 2.5 moles of KCl.

In 75 mL 0.1875 moles of KCl.

So, 0.1875 moles of KCl will react with 0.09375 moles of Ag₂SO₄.

1.4 M Ag₂SO₄

1.4 moles of Ag₂SO₄ in 1000 mL

0.09375 moles of Ag₂SO₄ in 66.96 mL

B.

143.32 g of AgCl = 1 mole

72.8 g of AgCl = 0.5 mole

So, 0.5 mole of AgCl formed from 0.5 mole of KCl.

0.250 M KCl is

0.250 moles of KCl in 1000 mL

0.5 moles of KCl in 2000 mL

0.5 moles of KCl in 2 L.

C.

0.762 M KCl

In 1000 mL= 0.762 moles KCl

In 94.1 mL = 0.0717 moles of KCl.

So, 0.0717 moles of KCl reacts with 0.03585 moles of Ag₂SO₄.

1 mole of Ag₂SO₄ in 1000 mL = 1 M

0.03585 moles of Ag₂SO₄ in 1000 mL = 0.03585 M

0.03585 moles of Ag₂SO₄ in 61.8 mL = (0.03585 x 61.8)/1000 M = 2.21 x 10^-3 M.

D.

0.236 M Ag2SO4

In 1000 mL 0.236 moles of Ag2SO4.

In 62.8 mL 0.0148 moles of Ag2SO4.

So 0.0148 moles of Ag2SO4 will react with 0.0296 moles of KCl.

1 mole of KCl in 1000 mL = 1 M

0.0296 moles of KCl in 1000 mL = 0.0296 M

0.0296 moles of KCl in 23.7 mL = (0.0296 x 23.7)/1000 M = 7.01 x 10^-4 M.