Concentration Of Ions In Solution
Concentration of Ions with Examples
Concentration of Ions with Examples
We examine concentration of ions with examples.
Instance: 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) + 2NOiii -(aq)
i mole Ca(NO3)2 gives one mole Ca+2 and ii moles NO3 - ions to solution.
1 mole Ca(NO3)2 gives one mole Ca+ii ion
0,2 mole Ca(NO3)2 gives ? mole Ca+ii ion
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?=0,2 mole Ca+2 ion
i mole Ca(NO3)ii gives 2 mole NOiii - ion
0,2 mole Ca(NO three)2 gives ? mole NOthree - ion
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?=0,4 mole NO3 - ion
Since volume of solution is 500 mL=0,five 50, molar concentration of solution becomes;
M=nsolution/V
M=0,two/0,5=0,iv mol/L
Molar concentrations of ions ;
[Ca+ii]=nCa+2/5=0,2/0,5=0,4 mol/Fifty
[NOthree -]=nNO3-/V=0,4/0,5=0,8 mol/L
Example: two,68 g Na2And thenfour.xH2O solute dissolves in h2o and 100 mL solution is prepared. If the concentration of Na+ ion in this solution is 0,2 molar, discover 10 in the formula of chemical compound. (Na2And then4=142 and HiiO=18)
Solution:
We kickoff find moles of Na+ ion using following concentration formula;
[Na+]=nNa+/5
Five=100mL=0,1L and [Na+]=0,2 molar
due northNa+=[Na+].Five=(0,1).(0,2)=0,02 mole
We observe mole of solution including 0,02 mole Na+;
1 mole NatwoSO4.xH2O includes 2 mole Na+ ion
? mole NatwoSOfour.xHtwoO includes 0,02 mole Na+ ion
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?=0,01 mole NatwoAnd so4.xH2O
Tooth mass of compound;
0,01 mole Na2And so4.xHiiO is two,68 g
1 mole Na2Soiv.xH2O is ? g
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?=268 g
NaiiAnd so4.xH2O=268 m
142 + ten(18)=268
ten=7
Example: We mix ii solutions having iv liters 0,two molar Kii(SO4) and 1 liter Altwo(And thenfour)3. If tooth concentration of And sofour -2 ion is 0,four tooth, find tooth concentration of Al2(SO4)3.
Solution:
Moles of Chiliadtwo(Then4)
northK2(SO4)=Five.M=four.0,2=0,eight mole
Since 1 mole K2(Then4) gives 1 mole SOfour -ii, 0,eight mole Thou2(SOfour) gives 0,8 mole And so4 -2
Moles of Al2(SO4)3
nAl2(SO4)3=Five.M=1.X=ten moles (x is molarity of Altwo(Theniv)3
Since 1 mole Alii(SOfour)3 gives three moles SOiv -2, 10 mole Alii(Sofour)3 gives 3x mole And then4 -ii
Full number of moles of SOiv -2 in solution is;
nSOiv -2=0,8 + 3x
Volume of solution is;
Vsolution =4 + 1=5 L
Tooth concentration of And sofour -2;
[SO4 -2]=nSOfour -ii/5sol
0,4=(0,8+3x)/5
x=0,4 molar.
Example: During dissolution of Al(NOiii)3 and Ca(NO3)2 in h2o, graph given below shows alter in the number of moles of Al+3 and NOiii - ions. If final ion concentration of Ca+2 is 0,05 molar, discover volume of solution.
Solution:
We encounter that final moles of NO3 - is 0,xvi and Al+iii is 0,04.
1 mole Al(NO 3)3 gives 1 mole Al+3 and iii mole NO3 -
If ane mole Al(NO 3)3 gives iii mole NO3 -
0,04 mole Al(NO 3)iii gives ? mole NOthree -
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?=0,12 mole NOthree -
Since there are 0,xvi mole NO3 -in solution, 0,xvi-0,12=0,04 mole NO3 -comes from Ca(NOiii)2.
mole Ca(NOthree)2 gives i mole Ca+2 and 2 mole NO3 -
If 1 mole Ca+2 reacts with 2 mole NOiii -
? mole Ca+two reacts with 0,04 mole NOthree -
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?=0,02 mole Ca+2
Tooth concentration of Ca+ii;
[Ca+2]=nCa+2/V
0,05=0,02/5
5=0,4 50=400 mL.
Solutions Exams and Problem Solutions
Concentration Of Ions In Solution,
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