Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

6b. Reacting mass chemical calculations - reacting masses, equations, volumes and concentrations - volumetric titrations (BUT method NOT using moles or molarity)

This page describes and explains, with fully worked out examples, how to do titration calculations just using concentrations measured in mass/volume units. This calculation method also involves reacting masses deduced from the balanced symbol equation.

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6b. Reacting masses, concentration of solution and volumetric titration calculations

NOT using the mole concept BUT see also

Basic acid-alkali titration questions using moles far more appropriate!

Advanced Level acid-alkali volumetric titration calculations

Concentration of a solution

For quantitative chemistry, it is important to know the concentration of solutions and be able to do calculations based on experimental results for chemical analysis.

The concentration of a solution is often measured in grams per decimetre cubed

so the units are expressed as gdm^-3, or g/dm³, which was g/litre!).

concentration = mass / volume (c = m / V)

concentration in gdm^-3, mass in g, volume in dm³

Its really important remember that 1 dm³ = 1000 cm³ or 1000 ml

and cm³/1000 = dm³

rearrangements of the formula using the triangle to help gives

mass = concentration x volume (m = c x V)

and volume = mass / concentration (V = m / c)

In the example questions I have used the shorthand formulae c = m / V, m = c x V and V = m / c

and RFM as an abbreviation of relative formula mass or molecular mass, and other shorthands:-

sf means numbers rounded to 2/3/4 significant figures and dp means round to 2/3/4 decimal places

Example 6b.1

(a) What is the concentration of a salt solution if you dissolve 10g of sodium chloride in 250 cm³ of water?

250 cm³ is equal to 250/1000 = 0.25dm³

therefore the concentration c = m / V = 10/0.25 = 40 g/dm³

(40g/litre in old units, still in common use!)

(b) What mass of the salt is required to make 200 cm³ of concentration 15g/dm³?

V = 200/1000 = 0.2 dm³

m = c x V = 15 x 0.2 =  3.0 g

(c) If you were given 8.0 g of salt, what volume of water, in dm³ and cm³, should you dissolve it in, to give a salt solution of concentration of 5g/dm³?

V = m / c = 8 / 5 = 1.6 dm³

V = 1.6 x 1000 = 1600 cm³

Examples 6b.2, 3 & 4 Quantitative calculations involving titrations

The next three examples are more complicated and involve using reacting mass ratio calculations which have been covered in detail in section 6a. Reacting mass ratio chemical calculations (not using moles)

Example 6b.2

0.5g of sodium chloride was dissolved in water and you are given a silver nitrate solution of concentration 30g/dm³. When you mix the two solutions you get a white precipitate of silver chloride.

Given the equation and the RFMs calculated from the atomic masses

Na = 23, Cl = 35.5, Ag = 107.8, O = 16, N = 14

Using an accurate burette, precisely what volume (in cm³) of the silver nitrate solution must be added to the sodium chloride solution in order to precipitate the maximum amount of silver chloride without wasting a drop of quite a costly solution! Three steps to the calculation:- (a) calculate the relative formula masses of the reactants, (b) the mass of silver nitrate reacting and (c) the volume of silver nitrate needed.

(b) From the equation calculate the mass of silver nitrate that reacts with 0.5 g of sodium chloride.

According to the symbol equation one RFM or 'molecule' of sodium chloride reacts with one RFM or 'molecule' of silver nitrate.

NaCl	:	AgNO3
58.5	:	169.8
0.5g	:	x g

Solving the ratio gives x = 0.5 x 169.8 / 58.5 = 1.451g of silver nitrate

(b) From your answer to (a) calculate the volume of silver nitrate needed.

V = m / c = 1.451 / 30 = 0.0484 dm³

V = 1000 x 0.0484 = 48.4 cm³

Example 6b.3

The reaction between hydrochloric acid and sodium hydroxide is

According to the symbol equation one RFM or 'molecule' of hydrochloric acid reacts with one RFM or 'molecule' of sodium hydroxide.

(The atomic masses involved are H = 1, Cl = 35.5, Na = 23 and O = 16, check the RFM for yourself)

Therefore from the relative formula masses (RFMs) 36.5g of HCl reacts with 40g of sodium hydroxide.

You are given a solution of hydrochloric acid of 7.3 g/dm³ (of HCl). 25.0 cm³ of sodium hydroxide solution was pipetted into a conical flask. On titration with the acid solution using a burette and suitable indicator, it was found that 14.6 cm³ of the acid solution was required to completely neutralise the alkaline sodium hydroxide. Calculate the concentration of the sodium hydroxide solution in g/dm³ via the stages outlined below

(a) The mass of hydrochloric acid reacting.

V = 14.6 cm³/1000 = 0.0146 dm³

m = c x V = 7.3 x 0.0146 =  0.1066 g HCl

(b) Calculate the mass of NaOH that reacts with 0.1066 g of HCl

HCl	:	NaOH
36.5	:	40
0.1066g	:	x g

Solving the ratio x = 0.1066 x 40/36.5 =  0.1168 g

(c) Calculate the concentration of NaOH

V = 25 cm3/1000 = 0.025 dm3

c = 0.1168/0.025 =  4.67 g/dm³ (2dp, 3sf)

Now it is possible to titrate the acid solution with the alkali solution (or vice versa) to obtain an unknown concentration of one of the solutions. One concentration must be known and the two volumes (acid and alkali) which react together exactly i.e. no excess of either reactant solution. How to do this is described in Section 12. acid-alkali titrations

Example 6b.4

Citric acid is the most common acid in citrus fruits and can be estimated by titration with sodium hydroxide.

citric acid	+	sodium hydroxide	===>	sodium citrate	+	water
(aq)	+	3NaOH(aq)	===>	(aq)	+	H2O(l)
RFM = 192 for C6H8O7		RFM = 40 but reacting mass is 3 x 40 = 120

You are a given of sodium hydroxide with a concentration of 12.0 g/dm³.

10 cm³ of squeezed lemon juice was pipetted into a flask. Using a burette and suitable indicator the lemon juice required a titration volume of 35.5 cm³ of the sodium hydroxide solution to completely neutralise it.

Assuming all the acidity is due to citric acid, calculate the concentration of citric acid in the lemon juice.

(a) What mass of sodium hydroxide reacted with the lemon juice?

V = 35.5 cm³/1000 = 0.0355 dm³

m = c x V = 12.0 x 0.0355 = 0.426g NaOH

(b) What mass of citric acid reacted with the 0.426 g of sodium hydroxide?

C6H8O7	:	3NaOH
192	:	3 x 40 = 120
x g	:	0.426 g

Solving the ratio x = 0.426 x 192/120 = 0.6816 g C₆H₈O₇

(c) Calculate the concentration of citric acid in the lemon juice?

V = 10 cm3/1000 = 0.01 dm3

c = m / V = 0.6816/0.01 = 68.2 g/dm³ C₆H₈O₇

See also section 11. for molarity calculations and Section 12. volumetric titration questions using moles

QUESTIONS answers at the end so you can selectively printout the questions and return for the answers!

• Question 6b.1:  A potassium sulfate solution A has a concentration of 6.5g/dm³

  • (a) What mass of potassium sulfate is present in 25cm³ of solution A?

  • (b) What volume of solution A, in cm³, contains 3.0g of potassium sulphate?

  • (c) What mass of the salt is required to make 250 cm³ of a solution B so it contains 3.75 g/dm³ of potassium sulfate?

  • (d) If 12.5g of potassium sulphate was dissolved in 1500 cm3 of water to make up solution C, what is the concentration of the salt?

• Question 6b.2: Nitric acid - sodium hydroxide titration

  • The reaction equation for the neutralisation of nitric acid by sodium hydroxide to form sodium nitrate is

  • HNO₃(aq) + NaOH(aq) ===> NaNO₃(aq) + H₂O(l)

  • (a) Calculate the relative formula masses of nitric acid, sodium hydroxide and sodium nitrate

    • Atomic masses: H = 1, N = 14, O = 16, Na = 23

  • 25.0 cm³ of a nitric acid solution was pipetted into a conical flask and titrated with a sodium hydroxide solution of concentration 5.0g/dm³. If it took 16.8 cm³ of the alkali to neutralise the acid, calculate the following

  • (b) The mass of sodium hydroxide that reacted with the nitric acid.

  • (c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.

  • (d) What was the concentration of the nitric acid in g/dm³?

  • (e) If the resulting neutral solution was carefully evaporated to dryness, what mass of sodium nitrate salt crystals would be left as a residue?

• Question 6b.3: Limewater analysis

  • The symbol equation for the neutralisation reaction between calcium hydroxide solution (limewater) and hydrochloric acid to form calcium chloride and water is

  • Ca(OH)₂(aq) + 2HCl(aq) ===> CaCl₂(aq) + 2H₂O(l)

  • 50 cm³ of a limewater solution was completely neutralised by 9.7 cm³ of a hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g HCl/dm³ calculate the concentration of calcium hydroxide in the water by the following method.

  • (a) calculate the relative formula masses of the reactants.

    • Atomic masses: Ca = 40, O = 16, H = 1, Cl = 35.5

  • (b) What mass of hydrochloric acid (as HCl) reacted with the limewater?

  • (c) What mass of calcium hydroxide reacted with the hydrochloric acid?

  • (d) What is the concentration of calcium hydroxide in g/dm³?

• Question 6b.4: Sulfuric acid - potassium hydroxide reaction to make potassium sulfate

  • The neutralisation reaction between sulphuric acid and potassium hydroxide is

  • H₂SO₄(aq) + 2KOH(aq) ===> K₂SO₄(aq) + 2H₂O(l)

  • A solution of potassium hydroxide contained 100g/dm³.

  • A solution of sulfuric acid contained 120g/dm³.

  • What volume in cm³, of the sulfuric acid solution, is required to neutralise 250 cm³ of the potassium hydroxide solution?

  • Atomic masses: H = 1, S = 32, O = 16, K = 39

ANSWERS

• Question 6b.1:  A potassium sulfate solution has a concentration of 6.5g/dm³

  • (a) What mass of potassium sulfate is present in 25cm³ of the solution?

    • 25cm³ = 25/1000 = 0.025 dm³

    • m = c x V

    • mass = 6.5 x 0.025 =  0.1625 g

  • (b) What volume of solution, in cm³, contains 3.0g of potassium sulphate?

    • V = m / c = 3.0 / 6.5 =  0.462 dm³ (to 3 sf)

    • V = 1000 x 0.462 = 462 cm³

  • (c) What mass of the salt is required to make 250 cm³ of a solution B so it contains 3.75 g/dm³ of potassium sulfate?

    • volume = 250/1000 = 0.25 dm³

    • m = c x V = 3.75 x 0.25 = 0.9375 g

  • (d) If 12.5g of potassium sulphate was dissolved in 1500 cm³ of water to make up solution C, what is the concentration of the salt?

    • V = 1500 cm³/1000 = 1.5 dm³

    • c = m / V = 12.5/1.5 = 8.33 g/dm³

• Question 6b.2: Nitric acid - sodium hydroxide titration

  • (a)

    nitric acid	+	sodium hydroxide	===>	sodium nitrate	+	water
    HNO3(aq)	+	NaOH(aq)	===>	NaNO3(aq)	+	2H2O(l)
    RFM 63		40		85		-

  • 25.0 cm³ of a nitric acid solution was pipetted into a conical flask and titrated with a sodium hydroxide solution of concentration 5.0g/dm³. If it took 16.8 cm³ of the alkali to neutralise the acid, calculate the following

  • (b) The mass of sodium hydroxide that reacted with the nitric acid.

    • V = 16.8 cm³/1000 = 0.0168 dm³

    • m = c x V = 5.0 x 0.0168 =  0.084g NaOH

  • (c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.

    • HNO3	:	NaOH
      63	:	40
      x g	:	0.084 g

    • Solving the ratio: x = 0.084 x 63/40 = 0.1323 g HNO₃

  • (d) What was the concentration of the nitric acid in g/dm³?

    • V = 25.0 cm³/1000 = 0.025 dm³

    • c = m / V = 0.1323 / 0.025 = 5.29 g/dm³ (3sf, 2dp)

  • (e) If the resulting neutral solution was carefully evaporated to dryness, what mass of sodium nitrate salt crystals would be left as a residue?

    • NaOH	:	NaNO3
      40	:	85
      0.084 g	:	x g

    • Solving the ratio: x = 0.084 x 85/40 = 0.1785 g NaNO₃
    • You can also use HNO₃ instead of NaOH i.e. 63 : 85, it doesn't matter which you use.

• Question 6b.3: Limewater analysis

  • The symbol equation for the neutralisation reaction between calcium hydroxide solution (limewater) and hydrochloric acid to form calcium chloride and water is

  • 50 cm³ of a limewater solution was completely neutralised by 9.7 cm³ of a hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g HCl/dm³ calculate the concentration of calcium hydroxide in the water by the following method.

  • (a) Calculate the relative formula masses of the reactants.

  • calcium hydroxide	+	hydrochloric acid	===>	calcium chloride	+	water
    Ca(OH)2(aq)	+	2HCl(aq)	===>	CaCl2(aq)	+	2H2O(l)
    RFM 74		36.5		-

  • (b) What mass of hydrochloric acid (as HCl) reacted with the limewater?

    • V = 9.7 cm³/1000 = 0.0097 dm³

    • m = c x V = 3.65 x 0.0097 =  0.0354 g HCl

  • (c) What mass of calcium hydroxide reacted with the hydrochloric acid?

    • Ca(OH)2	:	2HCl
      74	:	2 x 36.5 = 73
      x g	:	0.0354 g

    • Solving the ratio: x = 0.0354 x 74/73 = 0.03588 g Ca(OH)₂

  • (d) What is the concentration of calcium hydroxide in g/dm³?

    • V = 50 cm³/1000 = 0.05 dm³

    • c = m / V = 0.03588/0.05 = 0.718 g/dm³ Ca(OH)₂

• Question 6b.4: Sulfuric acid - potassium hydroxide reaction to make potassium sulfate

  • H₂SO₄(aq) + 2KOH(aq) ===> K₂SO₄(aq) + 2H₂O(l)

  • Potassium hydroxide contained 100g/dm³ and sulfuric acid contained 120g/dm³.

  • What volume in cm³, of the sulfuric acid solution, is required to neutralise 250 cm³ of the potassium hydroxide solution?

  • To calculate mass of KOH

    • V = 250 cm3/1000 = 0.25 dm³

    • m = c x V = 100 x 0.25 =  25g of KOH

  • To calculate mass of H₂SO₄ needed

    • H2SO4	:	2KOH
      98	:	2 x 56 = 112
      x g	:	25 g

    • Solving the ratio: x = 25 x 98/112 = 21.875 g H₂SO₄

  • To calculate the volume of sulphuric acid needed.

    • V = m / c = 21.875/120 =  0.1823 dm³

    •  0.1823 x 1000 =  182 cm³ (3sf, nearest cm³)

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