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Concentration Calculations (SL IB Chemistry)

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Philippa Platt

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Concentration Calculations

Titrations

  • Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a standard solution) to determine the concentration of another unknown solution
  • The technique most commonly used is a titration
  • The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette
  • The steps in a titration are:
    • Measuring a known volume (usually 20 or 25 cm3) of one of the solutions with a volumetric or graduated pipette and placing it into a conical flask
    • The other solution is placed in the burette
    • A few drops of the indicator are added
    • The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator just changes colour
    • Multiple trials are carried out until concordant results are obtained

Calculating concentration

  • Concentration calculations involve bringing together the skills and knowledge you have acquired in molar concentration and applying them to problem solving
  • You should be able to easily convert between moles, mass, concentrations and volumes ( of solutions and gases)
  • The four steps involved in problem solving are:
    • write the balanced equation for the reaction
    • determine the mass/ moles/ concentration/ volume of the of the substance(s) you know about
    • use the balanced equation to deduce the mole ratios of the substances present
    • calculate the mass/ moles/ concentration/ volume of the of the unknown substance(s)

Worked example

25.0 cm3 of 0.050 mol dm-3 sodium carbonate was completely neutralised by 20.0 cm3 of dilute hydrochloric acid.Calculate the concentration in mol dm-3 of the hydrochloric acid.

Answer:

Step 1: Write the balanced equation for the reaction

Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

Step 2: Determine the moles of the known substance, in this case sodium carbonate. Don't forget to divide the volume by 1000 to convert cm3 to dm3

moles = volume x concentration

amount (Na2CO3) = 0.0250 dm3 x 0.050 mol dm-3 = 0.00125 mol

Step 3: Use the balanced equation to deduce the mole ratio of sodium carbonate to hydrochloric acid:

1 mol of Na2CO3 reacts with 2 mol of HCl, so the mole ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 4: Calculate the concentration of the unknown substance, hydrochloric acid

concentration space equals space moles over volume concentration space open parentheses HCl close parentheses space equals space fraction numerator 0.00250 space mol over denominator 0.0200 space dm cubed end fraction equals space 0.125 space mol space dm to the power of negative 3 end exponent

Worked example

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react completely with 2.5 g of calcium carbonate.

Answer:

Step 1: Write the balanced equation for the reaction

CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

Step 2: Determine the moles of the known substance, calcium carbonate

Amount of CaCO3 = fraction numerator 2.5 space straight g over denominator 100.09 space straight g space mol to the power of negative 1 end exponent end fraction equals space 0.025 space mol

Step 3: Use the balanced equation to deduce the mole ratio of calcium carbonate to hydrochloric acid:

1 mol of CaCO3 requires 2 mol of HCl

So 0.025 mol of CaCO3 requires 0.050 mol of HCl

Step 4: Calculate the volume of HCl required

Volume of HCl = moles over concentration equals space fraction numerator 0.050 space mol over denominator 1.0 space mol space dm to the power of negative 3 end exponent end fraction equals space 0.050 space dm cubed

Examiner Tip

When performing titration calculations using monoprotic acids (meaning one H+) such as HCl, the number of moles of the acid and alkali will be the same. This allows you to use the relationship

C1V1 =C2V2

where C1 and V1 are the concentration and volume of the acid and C2 and V2 are the concentration and volume of the alkali. There is no need to convert the units of volume to dm3 as this is a ratio.Simply re-arrange the formula to solve for the unknown quantity.

Worked example

A 0.675 g sample of a solid acid, HA, was dissolved in distilled water and made up to 100.0 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against 0.100 mol dm-3 NaOH solution and 12.1 cm3 were required for complete reaction. Determine the molar mass of the acid.

Answer:

Step 1: Write the equation for the reaction

HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)

Step 2: Calculate the number of moles of the NaOH

n(NaOH)sample = open parentheses fraction numerator 12.1 space cm cubed over denominator 1000 end fraction close parentheses space dm cubed cross times 0.100 space mol space dm to the power of negative 3 end exponent equals space 1.21 space cross times 10 to the power of negative 3 end exponent space mol

Step 3: Deduce the number of moles of the acid

Since the acid is monoprotic the number of moles of HA is also 1.21 x 10-3 mol

This is present in 25.0 cm3 of the solution

Step 4: Scale up to find the amount in the original solution

straight n open parentheses NaOH close parentheses subscript original equals space fraction numerator 1.21 space cross times 10 to the power of negative 3 end exponent space mol space cross times 100.0 space cm cubed over denominator 25.0 space cm cubed end fraction equals space 4.84 space cross times 10 to the power of negative 3 end exponent space mol

Step 5: Calculate the molar mass

moles space equals space fraction numerator mass over denominator molar space mass end fraction

molar space mass space equals mass over moles equals space fraction numerator 0.675 space straight g over denominator 4.84 space cross times 10 to the power of negative 3 end exponent end fraction equals space 139 space straight g space mol to the power of negative 1 end exponent

Back titration

  • A back titration is a common technique used to find the concentration or amount of an unknown substance indirectly
  • The principle is to carry out a reaction with the unknown substance and an excess of a further reactant such as an acid or an alkali
  • The excess reactant, after reaction, is then analysed by titration and the mole ratios are used to deduce the moles or concentration of the original substance being analysed

Worked example

The percentage by mass of calcium carbonate, CaCO3, in a sample of marble was determined by adding excess hydrochloric acid to ensure that all the calcium carbonate had reacted. The excess acid left was then titrated with aqueous sodium hydroxide. A student added 27.20 cm3 of 0.200 mol dm-3 HCl to 0.188 g of marble. The excess acid required 23.80 cm3 of 0.100 mol dm-3 NaOH for neutralisation. Calculate the percentage of calcium carbonate in the marble.

Answer:

Step 1: Write the equation for the titration reaction:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Step 2: Calculate the number of moles of the NaOH

n(NaOH) = 0.02380 dm3 x 0.100 mol dm-3 = 2.380 x 10-3 mol

Step 3: Deduce the number of moles of the excess acid

Since the reacting ratio is 1:1 the number of moles of HCl is also 2.380 x 10-3 mol

Step 4: Find the amount of HCl in the original solution and then the amount reacted

n(HCl)original =  0.02720 dm3 x 0.200 mol dm-3 = 5.440 x 10-3 mol

n(HCl)reacted =  5.440 x 10-3 mol – 2.380 x 10-3 mol = 3.060 x 10-3 mol

Step 5: Write the equation for the reaction with the calcium carbonate

2HCl (aq) + CaCO3 (s) →  CaCl2 (aq) + CO2 (g) + H2O (l)

Step 6: Deduce the number of moles of the calcium carbonate that reacted

Since the reacting ratio is 2:1 the number of moles of CaCO3 is (3.060 x 10-3 mol) ÷ 2

n(CaCO3) = 1.530 x 10-3 mol

Step 7: Calculate the mass of calcium carbonate in the sample of marble

mass = moles x molar mass = 1.530 x 10-3 mol x 100.09 g mol-1 = 0.1531g

Step 8: Calculate the percentage of calcium carbonate in the marble

Percentage of CaCO3 in marble = fraction numerator 0.1531 space cross times space 100 over denominator 0.188 end fraction equals space 81.5 percent sign

Examiner Tip

Rounding off when you take averages. When you have an average of burette readings that comes to three decimal places, e.g.(23.20 cm3 + 23.25 cm3) ÷ 2 = 23.225 cm3

You CANNOT show more than two decimal places because that would make the average more precise than the readings.

To manage this situation you need to follow a simple rule. If the last digit is between a 5 and 9 then you round up; if the digit is between 0 and 4 you round down. So in this case the value recorded would be 23.23 cm3

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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.