## Concentration of Ions with Examples

Concentration of Ions with Examples

We examine concentration of ions with examples.

Example: 500 mL solution includes 0,2 mole Ca(NO3)2. Find concentration of ions in this solution.

When Ca(NO3)2 dissolves in water;

Ca(NO3)2(aq) → Ca+2(aq) + 2NO3-(aq)

1 mole Ca(NO3)2 gives 1 mole Ca+2 and 2 moles NO3- ions to solution.

1 mole Ca(NO3)2 gives 1 mole Ca+2 ion

0,2 mole Ca(NO3)2 gives ? mole Ca+2 ion

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?=0,2 mole Ca+2 ion

1 mole Ca(NO3)2 gives 2 mole NO3- ion

0,2 mole Ca(NO3)2 gives ? mole NO3- ion

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?=0,4 mole NO3- ion

Since volume of solution is 500 mL=0,5 L, molar concentration of solution becomes;

M=nsolution/V

M=0,2/0,5=0,4 mol/L

Molar concentrations of ions ;

[Ca+2]=nCa+2/V=0,2/0,5=0,4 mol/L

[NO3-]=nNO3-/V=0,4/0,5=0,8 mol/L

Example: 2,68 g Na2SO4.xH2O solute dissolves in water and 100 mL solution is prepared. If the concentration of Na+ ion in this solution is 0,2 molar, find x in the formula of compound. (Na2SO4=142 and H2O=18)

Solution:

We first find moles of Na+ ion using following concentration formula;

[Na+]=nNa+/V

V=100mL=0,1L and [Na+]=0,2 molar

nNa+=[Na+].V=(0,1).(0,2)=0,02 mole

We find mole of solution including 0,02 mole Na+;

1 mole Na2SO4.xH2O includes  2 mole Na+ ion

? mole Na2SO4.xH2O includes  0,02 mole Na+ ion

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?=0,01 mole Na2SO4.xH2O

Molar mass of compound;

0,01 mole Na2SO4.xH2O is    2,68 g

1 mole  Na2SO4.xH2O is      ? g

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?=268 g

Na2SO4.xH2O=268 g

142 + x(18)=268

x=7

Example: We mix two solutions having 4 liters 0,2 molar K2(SO4) and 1 liter Al2(SO4)3. If molar concentration of SO4-2 ion is 0,4 molar, find molar concentration of Al2(SO4)3.

Solution:

Moles of K2(SO4)

nK2(SO4)=V.M=4.0,2=0,8 mole

Since 1 mole K2(SO4) gives 1 mole SO4-2, 0,8 mole K2(SO4) gives 0,8 mole SO4-2.

Moles of Al2(SO4)3,

nAl2(SO4)3=V.M=1.X=x moles (x is molarity of Al2(SO4)3

Since 1 mole Al2(SO4)3 gives 3 moles SO4-2, x mole Al2(SO4)3 gives 3x mole SO4-2

Total number of moles of SO4-2 in solution is;

nSO4-2=0,8 + 3x

Volume of solution is;

Vsolution=4 + 1=5 L

Molar concentration of SO4-2;

[SO4-2]=nSO4-2/Vsol

0,4=(0,8+3x)/5

x=0,4 molar.

Example: During dissolution of Al(NO3)3 and Ca(NO3)2 in water, graph given below shows change in the number of moles of Al+3 and NO3- ions. If final ion concentration of Ca+2 is 0,05 molar, find volume of solution.

Solution:

We see that final moles of NO3- is 0,16 and Al+3 is 0,04.

1 mole Al(NO3)3 gives 1 mole Al+3 and 3 mole  NO3-

If 1 mole Al(NO3)3 gives  3 mole  NO3-

0,04 mole Al(NO3)3 gives     ? mole NO3-

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?=0,12 mole NO3-

Since there are 0,16 mole NO3-in solution, 0,16-0,12=0,04 mole NO3- comes from Ca(NO3)2.

mole Ca(NO3)2 gives 1 mole Ca+2 and 2 mole  NO3-

If 1 mole Ca+2 reacts with  2 mole  NO3-

? mole Ca+2 reacts with  0,04 mole NO3-

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?=0,02 mole Ca+2

Molar concentration of Ca+2;

[Ca+2]=nCa+2/V

0,05=0,02/V

V=0,4 L=400 mL.

Solutions Exams and  Problem Solutions

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