a)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
Molarity(in M)=mass of solute(in g)volume of solution(in L)
a)
Answer to Problem 34E
0.0200 mole of Sodium phosphate in 10.0 mL of solution.
MNa+ 6.00 M MPO43- 2.00 M
Explanation of Solution
Record the given data
Moles of Sodium phosphate= 0.200 moles
Volume of the solution= 10.0 mL
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions,
Na3PO4(s)→3Na+(aq)+PO43-(aq)
The molarity of ions can be calculated by the formula,
Molarity=mass of solutevolume of solution
Therefore, by substituting the given info in the formula, the concentrations of separate ions can be given as,
Molarity=0.02000.0100=2.00M
In order, to find the concentration of the separate ion,
Concentration of individual ion= number of atoms×the calculated molarity
Therefore, the concentrations of separate ions can be given by,
3Na+=3×2.00=6.00 MNa+PO43- =1×3.00= 2.00 MPO43-
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual Na+ and PO43- are found to be 6.00 MNa+ and 2.00 MPO43- respectively.
b)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
Molarity(in M)=mass of solute(in g)volume of solution(in L)
b)
Answer to Problem 34E
0.300 mole of Barium nitrate in 600 mL of solution.
MBa2+ 0.500 M MNO3- 1.00 M
Explanation of Solution
Record the given data
Moles of Barium nitrate= 0.300 mole
Volume of the solution= 600 mL
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions
Ba(NO3)2(s)→Ba2++2NO32-
The molarity of ions can be calculated by the formula,
Molarity=mass of solutevolume of solution
Therefore, by substituting the given info in the formula, molarity can be calculated by,
Molarity=0.3000.600=0.500M
In order, to find the concentration of the separate ion,
Concentration of individual ion= number of atoms×the calculated molarity
Therefore, the concentrations of separate ions can be given by,
Ba2+=1×0.5=0.5 MBa2+NO3-= 2×0.5= 1.00 MNO3-
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual Ba2+ and NO3- are 0.5 MBa2+ and 1.00 MNO3- respectively.
c)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
Molarity(in M)=mass of solute(in g)volume of solution(in L)
c)
Answer to Problem 34E
1.00 g of Potassium chloride in 0.500 L of solution.
MK+ 0.0268 M MCl- 0.0268M
Explanation of Solution
Calculation
Record the given info
Mole of Potassium chloride=1.00 gram
Volume of solution= 0.500 L
for the concentration of individual ions is as follows,
The balanced equation can be given as,
KCl→K++Cl-
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Potassium chloride can be given as (1×40) + (1×35.5) =74.55 g/mol
Amount of Potassium chloride= MassMolar Mass
= 174.55=0.01341 mol
Concentration=amountvolume =0.013410.500=0.0268 M
In order, to find the concentration of the separate ion,
Concentration of individual ion= number of atoms×the calculated molarity
Therefore, the concentrations of separate ions can be given by,
K+= 1×0.0268=0.0268 MK+Cl- =1×0.0268= 0.0268 MCl-
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual K+ and Cl- are 0.0268 MK+ and 0.0268 MCl-
d)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
Molarity(in M)=mass of solute(in g)volume of solution(in L)
d)
Answer to Problem 34E
132 g of Ammonium sulphate in 1.50 L of solution.
MNH4+ 0.66 M MSO42- 1.32 M
Explanation of Solution
Record the given info
Mole of Ammonium sulphate=132 gram
Volume of solution=1.50L
Calculation for the concentration of individual ions is as follows,
The balanced equation for dissolving ions,
(NH4)2SO4→2NH42++SO42-
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Ammonium sulphate is 132 g/mol
Amount of Ammonium sulphate= MassMolar Mass
Amount of Ammonium sulphate =132132=1.00 mol
Therefore, from the amount of Ammonium sulphate, the molarity can be calculated as,
Concentration=amountvolume
Concentration=11.5=0.666 M
In order, to find the concentration of the separate ion,
Concentration of individual ion= number of atoms×the calculated molarity
Therefore, the concentrations of separate ions can be given by,
NH4+ =2×0.666= 1.32 MNH4+SO42-=1×0.666= 0.666 MSO42-
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual NH4+ and SO42- are
1.32 MNH4+ and 0.666 MSO42-.
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Chapter 4 Solutions
Lab Manual for Zumdahl/Zumdahl/DeCoste¿s Chemistry, 10th Edition
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