Ionic molarity

Chemistry Level 4

In a beaker containing 500 ml 500\text{ ml} of 0.12 M 0.12\text{ M} aqueous F e ( N O 3 ) X 3 \ce{Fe{(NO_3)}_3} , 100 ml 100\text{ ml} of aqueous 1 1 M F e Cl X 3 \ce{Fe{Cl}_3} is added, then find the molarity of the cations in the mixture.

Assume that the contents of the beaker do not react.


The answer is 1.067.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Apr 27, 2016

For ferric nitrate:-
0.12 = x 500 × 1000 0.12 = \dfrac{x}{500} × 1000 (Here x is the number of moles of ferric nitrate)
x = 0.06 x = 0.06
F e N O 3 X 3 → Fe X 3 + + 3 N O 3 X − \ce{Fe{NO_3}_3 -> {Fe}^{3+} + 3{NO_3}^{-}}
1 mole of ferric nitrate gives 1 mole of ferric ion.
So, 0.06 mole of ferric nitrate gives 0.06 moles of ferric ion.


For ferric chloride:-
1 = x 100 × 1000 1 = \dfrac{x}{100} × 1000 (Here x is the number of moles of ferric chloride)
x = 0.1 x = 0.1
F e Cl X 3 → Fe X 3 + + 3 Cl X − \ce{Fe{Cl}_3 -> {Fe}^{3+} + 3{Cl}^{-}}
1 mole of ferric chloride gives 1 mole of ferric ion.
So, 0.1 mole of ferric nitrate gives 0.1 moles of ferric ion.

So, total number of ferric ions = 0.1 + 0.06 = 0.16 0.1 + 0.06 = 0.16

So, molarity of ferric ions = 0.16 × 1000 150 = 1.067 \dfrac{0.16 × 1000}{150} = \boxed{1.067}

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