ways of expressing concentration
TRANSCRIPT
Ways of Expressing Concentration
Outline
Molarity
Analytical molarity (formality)
Equilibrium molarity
Molality
Percent concentration
% (w/w)
% (v/v)
% (w/v)
Parts per thousand (ppt), parts per million (ppm), and parts per billion (ppb)
Solution-diluent volume ratio
p-function
Normality
Titer
Density and specific gravity
Molarity, M
nsol' mL
solute mmol
nsol' L
solute molM
Analytical molarity (formality), Cx
nsol' L
solute mol totalCx
Analytical molarity, Cx
= Formality, F
When solute dissociates or ionizes completely in H2O formality is preferred to molarity in expressing concentration.
Cx describes the preparation of a solution of particular molarity.
Examples
1. Describe the preparation of NaOH sol’n with analytical concentration equal to1.00 M.
Examples
Answer:
Dissolve 1.00 mol NaOH (98.00 g NaOH) in H2O.
↓
Dilute to 1.00 L.
Examples
2. Calculate the molar concentration of ethanol in an aqueous solution that contains 2.30 g of C2H5OH (46.07 g/mol) in 3.50 L.
Examples
Answer:
0.0143 M
Examples
3. Calculate the analytical concentration of the solute species in aqueous solutions that contains 285 mg trichloroacetic acid, Cl3CCOOH (163.4 g/mol), in 10.0 mL.
Examples
Answer:
0.174 M
Equilibrium molarity, [X]
- molarity of a given species in a solution at equilibrium
Example
Calculate the equilibrium molar concentration of the solute species in aqueous solutions that contains 285 mg trichloroacetic acid, Cl3CCOOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized).
Example
Answer:
[Cl3CCOOH] = 0.047 M
[Cl3CCOO-] = 0.127 M
Molality, m
solvent kg
solute molm
Percent concentration
100%solution of weight
solute of weight%(w/w)
Percent concentration
100%solution of volume
solute of volume%(v/v)
Percent concentration
100%mL solution, of volume
g solute, of weight%(w/v)
Examples
1. What is the molality of 87.0% (w/w) H3PO4 (98.00 g/mol) solution?
Examples
Answer:
68.3 m
Examples
2. What is the molarity of 37.7% (w/w) HCl (36.46 g/mol) solution? The density of the solution is 1.19 g/mL.
Examples
Answer:
12.3 M
Parts per thousand (ppt)
1000solution mass
solute masscppt
Parts per million (ppm)
6ppm 10
solution mass
solute massc
1 ppm = 1 mg/L
Parts per billion (ppb)
9ppb 10
solution mass
solute massc
1 ppb = 1 μg/L
Examples
1. There is 0.050 mg of PCB in 2 700 kg of soil. What is the concentration in ppb of PCB in the sample?
Examples
Answer:
0.019 ppb
Examples
2. What is the molarity of K+ in aqueous solution that contains 63.3 ppm K3Fe(CN)6 (329.3 g/mol)?
Examples
Answer:
5.77 X 10-4 M
Solute-diluent volume ratios
solute: solvent
Example:
1:4 HCl solution = 1 volume of conc. HCl +
4 volumes of H2O
p-function
[X] -logpX
Example
The pAg of a solution is 9.00. Calculate the molar concentration of Ag+ in the solution.
Example
Answer:
1.000 X 10-9 M
Normality, N
nsol' L
solute equivalentN
solute EW
solute gsolute equivalent
n
MWEW
Determining n
n depends on the type of reaction:
1. Acid-base reaction
2. Oxidation-reduction reaction
3. Precipitate formation and complex formation
Acid-base reaction
Acid:
n = number of H+ a mole of the acid supplies
Base:
n = number of H+ a mole of the base reacts with
Examples
1. H3PO4 + NaOH → NaH2PO4 + H2O
H3PO4: n = 1
NaOH: n = 1
Examples
2. H3PO4 + 2NaOH → Na2HPO4 + 2H2O
H3PO4: n = 2
NaOH: n = 1
Oxidation-reduction reaction
Species:
n = number of moles of electrons a mole of the species gains or loses
Half-reaction or net reaction can be used in determining n.
Examples
1. MnO4- + C2O4
2- → Mn2+ + CO2
KMnO4: n = 5
Na2C2O4: n = 2
Examples
2. I2 + S2O32- → I- + S4O6
2-
I2: n = 1
S2O32-: n = 1
Precipitate and complex formation
cation:
n = charge of cation
anion:
n = charge of cation x coefficient of cation
coefficient of anion
Examples
1. Ag+ + Cl- AgCl(s)
AgNO3: n = 1
KCl: n = 1
Examples
2. Ba2+ + SO42- BaSO4(s)
Ba(NO3)2: n = 2
Na2SO4: n = 2
Examples
3. Ag+ + 2CN- Ag(CN)2-
AgNO3: n = 1
KCN: n = 1/2
Examples
4. 3Cu(NO3)2 + 2Na3PO4 Cu3(PO4)2(s) + 6NaNO3
Cu(NO3)2: n = 2
Na3PO4: n = 3
Relation between N and M
N = nM
Significance of normality
Consider titration of analyte A with titrant B
aA + bB → cC + dD.
At the endpoint,
eq A = eq B.
Hence,
NAVA=NBVB.
Examples
1. Calculate g of Na2CO3 needed to prepare 5.000 L of 0.100 N Na2CO3 (105.99 g/mol) from the primary standard solid, assuming the reaction is to be used for titrations in which the reaction is
CO32- + 2H+ → H2O + CO2
Examples
Answer:
26.50 Na2CO3
Examples
2. A 0.2121-g sample of pure Na2C2O4 (134.00 g/mol) was titrated with 43.31 mL KMnO4. What is the normality of KMnO4 solution. The reaction is
2MnO4- + 5C2O4
2- + 16H+ → 2Mn2+ + 10CO2 8H2O
Examples
Answer:
0.07309 N
Titer
- weight of pure substance (other than the solute) corresponding to 1 mL of solution
Corresponding means:will react exactly withis chemically equivalent tois contained inmay be obtained frommay be substituted for
Titer
unit: mg/mL
Relation between titer and N
Titer = Nsoln x EWpure substance
Examples
1. 1.00 mL HCl neutralizes 4.00 mg NaOH.
HCl + NaOH → NaCl + H2O
What is the NaOH titer of the HCl solution?
Examples
Answer:
Titer = 4.00 mg/mL
Examples
2. What is the NH3(17.00 g/mol) titer of a 0.120 N solution of HCl?
Examples
Answer:
Titer = 2.04 mg/mL
Examples
3. Calculate the normality of an HCl solution having a Na2CO3(105.99 g/mol) titer of 5.00 mg/mL.
Examples
Answer:
Titer = 0.0943 N
Density
volume
massdensity
Specific gravity
- ratio of the mass of a substance to the mass of an equal volume of H2O at 4°C