Concentration in mol dm⁻³ counts the moles in each cubic decimetre
Definition
Concentration
The mass or amount of a solute dissolved in a given volume of solution.
Definition
Mole
The amount of a substance that contains the Avogadro constant of particles, and which has a mass in grams equal to its relative particle mass.
- A concentration in mol dm−3\text{mol dm}^{-3}mol dm−3 states the amount of solute dissolved in each dm3\text{dm}^3dm3 of solution.
- The relationship between amount, volume and concentration is: concentration=amount of solute in molvolume in dm3\text{concentration} = \frac{\text{amount of solute in mol}}{\text{volume in dm}^3}concentration=volume in dm3amount of solute in mol
- Rearranged to give the amount present: amount in mol=concentration×volume in dm3\text{amount in mol} = \text{concentration} \times \text{volume in dm}^3amount in mol=concentration×volume in dm3
- Volumes measured in cm3\text{cm}^3cm3 are divided by 100010001000 to convert them to dm3\text{dm}^3dm3.
- A more concentrated solution holds more solute in the same volume.
Common Mistake
- Using a volume in cm3\text{cm}^3cm3 without converting makes the answer one thousand times too large.
- 25.0 cm325.0\ \text{cm}^325.0 cm3 is 0.0250 dm30.0250\ \text{dm}^30.0250 dm3, and that is the figure the equation takes.
Converting between g dm⁻³ and mol dm⁻³
Definition
Relative formula mass
The sum of the relative atomic masses of all the atoms shown in the formula of a substance, given the symbol Mr.
- A concentration in g dm−3\text{g dm}^{-3}g dm−3 states the mass of solute in each dm3\text{dm}^3dm3 of solution.
- Dividing by the relative formula mass converts a mass to an amount: concentration in mol dm−3=concentration in g dm−3Mr\text{concentration in mol dm}^{-3} = \frac{\text{concentration in g dm}^{-3}}{M_r}concentration in mol dm−3=Mrconcentration in g dm−3
- Multiplying reverses the conversion: concentration in g dm−3=concentration in mol dm−3×Mr\text{concentration in g dm}^{-3} = \text{concentration in mol dm}^{-3} \times M_rconcentration in g dm−3=concentration in mol dm−3×Mr
- Sodium hydroxide has an MrM_rMr of 404040, so 4.0 g dm−34.0\ \text{g dm}^{-3}4.0 g dm−3 is 0.10 mol dm−30.10\ \text{mol dm}^{-3}0.10 mol dm−3.
- The conversion factor is the MrM_rMr of the solute, so it changes from one solution to another.
Note
The same solution can be quoted either way, so checking the unit comes before any calculation.
A titration measures the volume that exactly reacts
Definition
Titration
A method that finds the exact volume of one solution that reacts with a measured volume of another.
Definition
Indicator
A substance that changes colour to show whether a solution is acidic, neutral or alkaline.
- A pipette delivers a fixed, accurately known volume of one solution into a conical flask.
- A burette adds the second solution a little at a time, with its volume read to 0.05 cm30.05\ \text{cm}^30.05 cm3.
- A few drops of a suitable indicator show when the reaction is exactly complete.
- The end point is the first permanent colour change, and the burette reading is taken there.
- The volume added is the titre, and concordant titres are averaged before any calculation.
Practical
- Method: pipette 25.0 cm325.0\ \text{cm}^325.0 cm3 of alkali into a conical flask, add a few drops of indicator, then add acid from a burette until the colour just changes.
- Rough then accurate: a first quick titration finds the approximate volume, and later runs are added dropwise near that point.
- Concordant results: titres within 0.10 cm30.10\ \text{cm}^30.10 cm3 of one another are averaged, and the rough titre is left out.
- Indicator choice: phenolphthalein or methyl orange gives a sharp change, while litmus changes too gradually.
A worked titration calculation
- 25.0 cm325.0\ \text{cm}^325.0 cm3 of sodium hydroxide of concentration 0.100 mol dm−30.100\ \text{mol dm}^{-3}0.100 mol dm−3 needs 20.0 cm320.0\ \text{cm}^320.0 cm3 of hydrochloric acid.
- First find the amount of the solution whose concentration is known: n(NaOH)=0.100×0.0250=2.50×10−3 moln(\text{NaOH}) = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\ \text{mol}n(NaOH)=0.100×0.0250=2.50×10−3 mol
- The balancing numbers in the equation are one to one, so the acid supplies the same amount.
- Then divide by the volume of acid, converted to dm3\text{dm}^3dm3: c(HCl)=2.50×10−30.0200=0.125 mol dm−3c(\text{HCl}) = \frac{2.50 \times 10^{-3}}{0.0200} = 0.125\ \text{mol dm}^{-3}c(HCl)=0.02002.50×10−3=0.125 mol dm−3
- The three steps are always the same: amount, ratio, then divide by the other volume.
Example
- Ratio that is not one to one: with H2SO4+2NaOH\text{H}_2\text{SO}_4 + 2\text{NaOH}H2SO4+2NaOH, the amount of acid is half the amount of alkali.
- The balancing numbers in the equation are what set that ratio, so the equation is written down first.
Finding an unknown volume instead
- The same three steps give a volume when the two concentrations are known.
- A titration of 25.0 cm325.0\ \text{cm}^325.0 cm3 of 0.100 mol dm−30.100\ \text{mol dm}^{-3}0.100 mol dm−3 sodium hydroxide against 0.200 mol dm−30.200\ \text{mol dm}^{-3}0.200 mol dm−3 acid gives n=2.50×10−3 moln = 2.50 \times 10^{-3}\ \text{mol}n=2.50×10−3 mol of each.
- Dividing the amount by the concentration gives the volume: V=2.50×10−30.200=0.0125 dm3V = \frac{2.50 \times 10^{-3}}{0.200} = 0.0125\ \text{dm}^3V=0.2002.50×10−3=0.0125 dm3
- Multiplying by 100010001000 converts the answer back to 12.5 cm312.5\ \text{cm}^312.5 cm3.
- A more concentrated acid needs a smaller volume, which is a useful check on the answer.
Exam technique
- Writing the balanced equation first fixes the ratio before any numbers are used.
- Every volume is converted to dm3\text{dm}^3dm3 at the start, so no factor of 100010001000 is left over.
- A final answer carries the unit, either mol dm−3\text{mol dm}^{-3}mol dm−3 or cm3\text{cm}^3cm3.
Self review
- What does a concentration of 0.50 mol dm−30.50\ \text{mol dm}^{-3}0.50 mol dm−3 tell you?
- How is a concentration in g dm−3\text{g dm}^{-3}g dm−3 converted to mol dm−3\text{mol dm}^{-3}mol dm−3?
- What is meant by the end point of a titration?
- 20.0 cm320.0\ \text{cm}^320.0 cm3 of 0.100 mol dm−30.100\ \text{mol dm}^{-3}0.100 mol dm−3 alkali reacts with 25.0 cm325.0\ \text{cm}^325.0 cm3 of acid in a one to one ratio. What is the concentration of the acid?
- Why must volumes be converted before the amount is calculated?
