L7 Solution Concentration
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Solution Concentration
Transcript of L7 Solution Concentration
Solution Concentration
Solution concentration…
…is a quantitative way of describing a solution.
…describes the amount of solute dissolved in a volume of solution.
Calculation concentration
c = amount of soluteamount of solution
There are many ways of calculating concentration, but all are essentially:
1. Percent weight by vol: % W/V
%W /V100 mL = m (of solute)
v (of solution)
• m must be in g• v must be in mL
Ex If 24 g of sodium nitrate is dissolved to
form 280 mL of solution, what is the percent concentration?
%W /V100 mL = mv
%W /V100 mL = 24 g
280 mL
%W/V = 8.5714… %W/V
%W/V = 8.6 %W/V
Ex What mass of lithium chloride is
dissolved to form 800 mL of a 12.8 %W/V solution?
%W /V100 mL = mv
12.8 %W /V100 mL = m
800 mL
m =102.4 g
m = 102 g
Ex When 14.4 g of salt is dissolved it forms
a 5.36% solution. Find the solution’s volume.
%W /V100 mL = mv
5.36 %W /V100 mL = 14.4 gv
v = 268.6567... mL
v = 269 mL
Ex Find the %W/V concentration when
450 mg of salt is dissolved to form 2.9 L of solution.
%W /V100 mL = mv
%W /V100 mL = 0.450 g
2900 mL
%W/V = 0.015517... %W/V
% = 0.016 %W/V
2. Percent vol by vol: % V/V
%V /V100 mL = v (of solute)
v (of solution)
• all v must be in mL
• Usually used with liquid solutes.
Ex A 750 mL bottle of wine has an alcohol
content of 11%. Find the volume of ethanol dissolved in the wine.
%V /V100 mL = vv
11 %V /V100 mL = v
750 mL
v = 82.5 mL
v = 83 mL
Ex What volume of 5.2% vinegear is
formed when 95 mL of acetic acid is dissolved?
%V /V100 mL = vv
5.2 %V /V100 mL = 95 mLv
v = 1826.923... mL
v = 1.8 x 103 mL or 1.8 L
3. Percent mass by mass: % W/W
%W /W100 g = m (of solute)
m (of solution)
• all m must be in g
• aka % weight by weight
Useful info:
For aqueous solutions: 1 g of solution = 1 mL 1 kg of solution = 1 L (unless stated otherwise in question)
Ex What mass of salt is dissolved in 6.0 L
of a 3.44% by mass solution?
%W /W100 g = mm
3.44 %W /W100 g = m
6000 g
m = 206.4 g
m = 0.21 kg
Note: in aqueous solution 6.0 L = 6.0 kg = 6000 g
4. Parts per million: ppm
ppm1 000 000 g =
m (of solute)m (of solution)
• all m must be in g
• generally used for dilute solutions
Ex Determine the ppm concentration of a
solution in which 3.49 g of toxic mercury ions are dissolved in 800 L of water.
ppm1 000 000 g =
mm
ppm1 000 000 g =
3.49 g800000 g
ppm = 4.3625 ppm
ppm = 4.36 ppm
In aqueous solution 800 L = 800 kg = 800000 g
Ex A 8.26 ppm solution of toxic PCBs
contain 0.12 g of the chemical. What is the solution volume?
ppm1 000 000 g =
mm
8.26 ppm1 000 000 g =
0.12 gm
m = 14527.845... g
m = 15 kg
In aqueous solution 15 kg = 15 L
5. Parts per billion: ppb
ppb1 000 000 000 g =
m (of solute)m (of solution)
• all m must be in g
• generally used for very dilute solutions
Ex Find the ppb concentration of a solution
in which 1.14 µg of toxic mercury ions are dissolved in 500 mL of water.
ppb1 000 000 000 g =
mm
ppb1 000 000 000 g =
1.14 x 10−6 g500 g
ppb = 2.28 ppb
In aqueous solution 500 mL = 500 g
Ex What mass of dioxin is dissolve in 35.2
L of a 48.2 ppb solution?
ppb1 000 000 000 g =
mm
48.2 ppb1 000 000 000 g =
m35200 g
m = 0.0016966… g
In aqueous solution 35.2 L = 35.2 kg = 35200 g
m = 1.70 mg
6. Parts per trillion: ppt
ppt1 000 000 000 000 g =
m (of solute)m (of solution)
• all m must be in g
• generally used for extremely dilute solutions
Ex If 3.8 µg of toxin is dissolved in a 67.4
ppt solution, find the solution volume?ppt
1 000 000 000 000 g =mm
67.4 ppt1 000 000 000 000 g =
3.8 x 10−6 gm
m = 56379.82... g
In aqueous solution 56 kg = 56 L
m = 56 kg
7. Amount (molar) concentration
most typical concentration used by chemists
if we simply say “concentration”, we mean amount (molar) concentration
C = nv
amount(mol)
volume(L)
Amount concentration(mol/L)
Other forms:
n=Cv v= nC
Ex. Find the amount concentration of a
solution when 16.7 g of sodium sulfate is dissolved to form 750 mL of solution.
C = nv
To find moles from mass need to use:
n = mM
=16.7 g
142.05 g /mol
n = 0.1175642.... mol
C = nv
=0.117564.... mol
0.750 L
= 0.15675231... mol/L
= 0.157 mol/L
Ex. Find the volume of a 0.234 mol/L
barium hydroxide solution if 5.8 g of solute was dissolved.
n = mM
n = 5.8 g171.35 g /mol
n = 0.033848847... mol
v = nC
v = 0.033848847... mol0.234 mol/L
v = 0.14465319... L
v = 0.14 L
Ex Find the solution’s concentration when
0.548 g of lead (II) nitrate is dissolved to form 950 mL of solution.
n = mM
= 0.548 g331.22 g /mol
= 0.0016544… mol
C = nv
= 0.001654... mol0.950 L
= 0.00174 mol/L
= 1.74 mmol/L
Your turn (Answers: Appendix A)
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