QUANTITATIVE CHEMISTRY
CHAPTER 4 QUANTITATIVE CHEMISTRYThe concentration of a solution is a quantitative expression of how much
solute is dissolved in the solution. Concentration can be expressed in g dm−3 or mol dm−3. Other units of concentration are used to describe very low
concentrations of solutions; these include parts per million, ppm (mg dm−3) and parts per billion, ppb (µg dm−3).
The concentration of a solution can be calculated by dividing the amount of solute (in mol) by the volume of the solution (in dm3).
c = n
V or n = cV
where c = concentration of solution (mol dm−3) n = number of mole of solute (mol) V = volume of solution (dm3)
This formula can be transposed to calculate any of the three variables, given the other two values.
Worked example 1
Calculate the concentration (in mol dm−3) of each of the following solutions.
a 2.0 mol of NaCl is dissolved in 8.0 dm3 of water.
c To solve this problem, we must fi rst calculate the number of moles of sucrose present.
It is now possible to calculate the molar concentration of glucose in the solution: Calculating concentration of a solution
1.5.2
Solve problems involving concentration, amount of solute and volume of solution.
© IBO 2007
CHEM COMPLEMENT
A drop in the ocean To help understand roughly what various concentrations mean, we can equate the concentration of a liquid dye in water as follows: 1% w/w equates to one drop in a teaspoon of water, 1 ppm to a drop in a full bathtub, 1 ppb to one drop in a full-sized swimming pool and 1 ppt (part per trillion) is a drop in 1000 swimming pools. It has been estimated that the world’s oceans may contain as much as US$10 trillion worth of gold. Unfortunately, this gold is found at a concentration of just 13 parts per trillion, or just 25 mg in an Olympic-sized swimming pool full of seawater. Two questions to consider: What is the current value of 25 mg of gold, and is it worth extracting the gold from seawater?
n
c V w
s
Figure 4.8.1 A formula triangle can be used to make it easier to rearrange this formula.
Worked example 2
Calculate the amount (in mol) of the substance stated that will be required to make the following solutions.
a 250 cm3 of 0.250 mol dm−3 H2SO4 b 10 cm3 of 0.825 mol dm3 NaOH Solution
a c = n
V, so n = cV where c = 0.250 mol dm−3, V = 0. 250 dm3, n = ? mol
∴ n (H2SO4 ) = 0.250 × 0.250 = 0.0625 mol b c = n
V, so n = cV where c = 0.825 mol dm−3, V = 0. 010 dm3,n = ? mol
∴ n(NaOH ) = 0.825 × 0.010 = 0.008 25 mol
It is important to recognize that the molar concentration refers to the amount of solute that has been dissolved in a given volume of the solvent water. For example, a 1.0 mol dm−3 solution of hydrochloric acid (HCl) would have 1 mole of HCl molecules per dm3 of water. As HCl ionizes fully in water, the HCl will exist as a 1.0 mol dm−3 solution of H+ ions and a 1.0 mol dm−3 solution of Cl− ions. Similarly, a 0.500 mol dm−3 solution of copper(II) sulfate would exist in solution as a 0.500 mol dm−3 solution of Cu2+ and a 0.500 mol dm−3 solution of SO42− ions.
How do we determine the concentration of individual ions in situations where the cations and anions are not in a 1:1 ratio? As we would expect, the concentration of individual ions in solution is in direct proportion to the chemical formula of the compound. Thus, in a 0.25 mol dm−3 solution of NiCl2, the concentration of nickel ions is 0.25 mol dm−3, while the chloride ion concentration is twice that, at 0.50 mol dm−3. A 2.00 mol dm−3 solution of ammonium sulfate ((NH4)2SO4) exists in solution as ammonium ions at a concentration of 4.00 mol dm−3 and sulfate ions as 2.00 mol dm−3.
If two or more solutions are mixed and there is no chemical reaction between individual solutes, we can calculate the concentration of each individual ion by determining the total amount of that ion present and dividing by the fi nal volume of solution.
Worked example 3
Aluminium sulfate is used in large quantities as a fl occulating agent in the treatment of wastewater. Calculate the molar concentration of both aluminium and sulfate ions when 40.00 kg of aluminium sulfate is dissolved in 250.0 dm3 of water.
Calculating concentrations of components of solutions
CHAPTER 4 QUANTITATIVE CHEMISTRY Solution
n(Al2(SO4)3) = m
M, where n = ?, m = 40 000 g, M(Al2(SO4)3) = 342.14 g mol−1
∴ n(Al2(SO4)3) = 40 000
342 14. = 116.9 mol
It is now possible to calculate the molar concentration of aluminium sulfate as a whole in the solution:
c = n
V, where c = ?, n = 116.9 mol, V = 250.0 dm3
∴ c(Al2(SO4)3) = 116 9 250 0 .
. = 0.4676 mol dm−3
∴ c(Al3+) = 2 × 0.4676 = 0.9352 M and c(SO42−) = 3 × 0.4676 = 1.403 mol dm−3
It is reasonably common for chemists to dilute a concentrated solution to produce a solution of a lower specifi ed concentration. As the original solution has merely been diluted with the addition of water, the actual amount (number of moles) of the original solute has not changed. The number of moles of solute in a solution can be determined by applying the formula n = cV. This leads us to a formula linking the concentration and volume of an original solution to its diluted form:
c1V1= c2V2
where c1= initial concentration of solution V1= initial volume of solution
c2= concentration of diluted solution V2= volume of diluted solution
Note that the formula is effectively a ratio and so units of volume must correspond; it is not necessary to convert all volumes to dm3 as long as they are the same.
Worked example 4
A student adds 200 cm3 of deionized water to 50 cm3 of 0.20 mol dm–3 HCl solution. Determine the concentration of the diluted hydrochloric acid solution.
Solution
Known Unknown
c1= 0.20 mol dm−3 c2= ?
V1= 50 cm3 V2= (200 + 50) = 250 cm3
c1V1= c2V2
∴ c2= c V V
1 1 2
= 0 20 50 250 . ×
= 0.040 mol dm−3
The concentration of the diluted solution is 0.040 mol dm−3. Dilution of solutions
Worked example 5
A solution of 20.0 cm3 of 0.250 mol dm−3 lead(II) nitrate is added to a large volume of 0.500 mol dm−3 potassium iodide solution to produce a brilliant yellow precipitate of lead(II) iodide. Write an equation for the reaction and so determine the mass of precipitate formed.
Solution
Note that in this problem we are informed that a ‘large volume’ of potassium iodide solution is used. This indicates that the potassium iodide must be in excess. The term ‘in excess’ implies that there is a suffi cient quantity of this reactant to ensure that all of the lead(II) nitrate solution is consumed in generating the precipitate.
Step 1: Write a balanced equation for the reaction.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Step 2: List all data given, including relevant units.
Known Unknown c(Pb(NO3)2) = 0.250 mol dm−3 m(PbI2) = ?
V(Pb(NO3)2) = 0.0200 dm3 M(PbI2) = 460.99 g mol−1 Step 3: Convert the data given for the ‘known’ to mol, using the relevant
formula.
c(Pb(NO3)2) = n V
∴ n(Pb(NO3)2) = cV = 0.250 × 0.0200 = 5.00 × 10−3 mol
Step 4: Use the chemical equation to determine the mole ratio of the unknown quantity to the known quantity.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) From the equation, n(PbI2) = 1
1 × n(Pb(NO3)2) = 5.00 × 10−3 mol Step 5: Convert this number of mole back into the relevant units of
the unknown.
n(PbI2) = m M
∴ m(PbI2) = 5.00 × 10−3 mol × 460.99 = 2.31 g The mass of lead(II) iodide precipitated is 2.31 g.
Worked example 6
Magnesium reacts readily with dilute hydrochloric acid to produce hydrogen gas, the presence of which can be ascertained by the ‘pop’ test.
a Calculate the volume of 1.50 mol dm−3 hydrochloric acid that would be required to completely consume 20.00 g of magnesium.
b Calculate the volume of hydrogen gas that would be produced in this reaction at STP (Vm= 22.4 dm3 mol−1).
Using concentrations and volumes of solution to find amounts of products
CHAPTER 4 QUANTITATIVE CHEMISTRY Solution
a Step 1: Write a balanced equation for the reaction.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Step 2: List all data given, including relevant units.
Known Unknown
m(Mg) = 20.00 g c(HCl) = 1.50 mol dm−3 M(Mg) = 24.31 g mol−1 V(HCl) = ?
Step 3: Convert the data given for the known quantity to mol, using the relevant formulas.
n(Mg) = m
M = 20 00 24 31
.
. = 0.8227 mol
Step 4: Use the chemical equation to determine the mole ratio of the unknown quantity to the known quantity.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation, n(HCl) = 2 × n(Mg) = 1.645 mol
Step 5: Convert this number of mole back into the relevant units of the unknown.
c(HCl) = n V
∴ V(HCl) = n
c = 1 645 1 50
.
. = 1.10
The volume of hydrochloric acid required is 1.10 dm3. b Step 1: Write a balanced equation for the reaction.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Step 2: List all data given, including relevant units.
Known Unknown
m(Mg) = 20.00 g Vm(H2) = 1.50 mol dm−3 M(Mg) = 24.31 g mol−1 V(H2) = ?
Step 3: Convert the data given for the ‘known’ to mol, using the relevant formula.
n(Mg) = m
M = 20 00 24 31
.
. = 0.8227 mol
Step 4: Use the chemical equation to determine the mole ratio of the unknown quantity to the known quantity.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation, n(H2) = n(Mg) = 0.8227 mol
Step 5: Convert this number of mole back into the relevant units of the unknown.
V(H2) = n × Vm
∴ V(H2) = 0.8227 × 22.4 = 18.4
The volume of hydrogen gas released is 18.4 dm3.
Figure 4.8.2 The reaction between magnesium and hydrochloric acid produces bubbles of hydrogen gas.
DEMO 4.3
Stoichiometry of a precipitation reaction
PRAC 4.6
Precipitation reactions PRAC 4.7
Chemical detectives
Worked example 7
Sulfuric acid is a strong acid that reacts completely with the alkali sodium hydroxide to produce water and the salt sodium sulfate. Calculate the volume of 0.486 mol dm−3 sodium hydroxide solution required to completely neutralize 35.5 cm3 of 0.606 mol dm−3 sulfuric acid.