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.