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How much 0. 5mna2so4 solution will completely precipitate the ba2 in 0. 7l of 0. 13mbacl2 solution?

Sagot :

The volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.

Calculations of number of moles for given molarity of solution

we use

Molarity =Moles of solute / volume of solution (in litre)

Now, for barium ions

Given,

Molarity of solution = 0.13 m

Volume of solution = 0.7L

By substituting all the value ,we have

Moles of Barium = 0.7 × 0.13

= 0.091

Chemical equation

BaSO4 » Ba2+ + So2-

By stoichiometry of equation

1 mole of sulphate ion precipitate 1 mole of barium ion

So, 0.091 mole of sulphate ion precipitate

0.091 of moles of Barium ion

Calculation of volume of sulfate ion

Now, calculate the volume of ions by using equation as given above

Mole of sulphate ion = 0.091 mol

Molarity concentration of sulphate ions = 0.05m

By substituting all value , we get

C = n / V

V = n/ C

= 0.091 / 0.5 = 0.04L

Thus, we calculated that the volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.

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