titration of a strong base-alkali with hydrochloric solution and titrating sulfuric acid with standardised solution of sodium hydroxide, method, procedure, apparatus, indicator, calculations Doc Brown's GCSE & A-level chemistry student exam revision notes for AQA, Edexcel, OCR, WJEC Eduqas & CCEA courses, US grades 9-10 AP Honors

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5. The titration of a strong base-alkali with hydrochloric solution and titrating sulfuric acid with standardised solution of sodium hydroxide

[Author © Dr Phil Brown GRIC, PhD: Doc Brown's chemistry exam revision notes on hydrochloric acid-sodium hydroxide titration suitable for students of UK GCSE level and Advanced A-level chemistry courses, ~US grades 9-12 chemistry notes [page updated RE-EDIT]

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Titration index * Full chemistry calculations index

Quiz 12 The basics of acid-alkali titration calculations Good exam practice questions for GCSE level

apparatus method diagram for titrating sulfuric acid or hydrochloric acid with standardised sodium hydroxide, titration calculation, phenophthalein indicator end-point

5. The titration of a strong base-alkali with hydrochloric solution

e.g titrating sodium hydroxide solution (pipetted) with standardised hydrochloric solution (of known concentration in burette) using phenolphthalein indicator

The apparatus, chemicals and indicator colours are illustrated in the diagram on the right.

Initially the burette is clamped carefully in position and filled with standard hydrochloric acid (e.g. 0.10 to 1.0 mol/dm³, but accurately known, preferably to 4 sig. figs.).

Wear safety glasses and initially work below eye level - so nothing can fall on your face!

Using a funnel, the hydrochloric acid is run through until the reading below the meniscus is 0.00 cm³ (the reading in the diagram is 7.00 cm³, which could represent a titration value). The burette is usually calibrated on the scale to 50.00 cm³ in 0.10 cm³ increments (only 10.00 cm³ in diagram - couldn't fit rest of scale on!)

The alkali solution is accurately measured out into the conical flask with e.g. a 25 cm³ pipette and suction bulb (see diagram further down).

Add a few drops of phenolphthalein indicator to the alkali solution and it should turn deep pink for an alkali. Carefully place the conical flask under the tip of the burette so drops don't go astray!

The titration: You carefully add small portions of the acid, swirling after each addition and checking the colour of the indicator (not shown in the diagram, but its good to stand the flask on white tile).

At the start of the titration the phenolphthalein indicator is a deep reddish-pink in alkaline solution.

As you add the acid you get 'splurges' of colourless solution until the mixture is swirled in the conical flask.

The swirling of the flask contents is important, it ensures all the added hydrochloric acid reacts with the sodium hydroxide.

Try to add dropwise when you seem to be near the faintest pinkness left in the solution near the endpoint.

The end-point is when the last trace of pink colour first disappears from the solution.

You taking the reading of the volume of the hydrochloric acid on the scale line the meniscus lies on (see lower part of diagram above on the right).

If you 'overshoot' the titration with excess acid, it still stays colourless and the result is invalid.
The titration should be repeated several times with other 25 cm³ portions and the average (mean) titration value calculated to use in any subsequent calculations.

How to calculate the concentration of the alkali is explained in the top half of the page.
The theory of which indicator to use is explained on the Changes in pH in a neutralisation, choice and use of indicators page.
There is also a section on errors and reasons for repeating a titration several times.

NOTE: If you have a question based on nitric acid (HNO₃), they are the same volumetric calculations as for hydrochloric acid (HCl) because both acids have one acidic proton (H), so balancing equations involves the same numbers.

Titration Calculation Example 12.3 analysing hydrochloric acid solution

Given the equation: NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)
This is read as 1 mol + 1 mol reactants ===> 1 mol + 1 mol products
25.00 cm³ portions of a dilute hydrochloric acid solution were titrated with a standard solution of sodium hydroxide of concentration 0.250 mol/dm³ (mol dm^-3).
See 2. The basic procedure for carrying out an acid - alkali titration
Using phenolphthalein indicator for the titration, it was found that the average titration was 18.50 cm³ of sodium hydroxide, calculate (i) the molarity of the hydrochloric acid and (ii) its concentration in g/dm³.
- moles NaOH = molarity NaOH x volume of NaOH (in dm³ = cm³/1000)
- moles NaOH = 0.250 x (18.5 / 1000) = 0.004625
- In the equation 1 mole of HCl reacts with 1 mole of HCl
- therefore in the titration reaction: moles HCl = moles NaOH
- therefore there were 0.004625 moles HCl in 25.00 cm³.
- molarity = moles / volume in dm³ (1 dm³ = 1000 cm³)
- (i) molarity HCl = 0.004625 / (25.00/1000) = 0.004625/0.025 = 0.185 mol/dm³
- (ii) concentration = molarity x formula mass
  - formula mass HCl = 1 + 35.5 = 36.5
  - = 0.185 x 36.5 = 6.75 g/dm³

Titration calculation Example 12.1 analysing sodium hydroxide solution
- Given the equation: NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)
  - This is read as 1 mol + 1 mol reactants ===> 1 mol + 1 mol products
- 25.0 cm³ of a sodium hydroxide solution was pipetted into a conical flask and titrated with a standard solution of 0.200 mol dm^-3 (0.2M) hydrochloric acid (mol dm^-3 means mol/dm³).
- Using phenolphthalein indicator for the titration it was found that 15.0 cm³ of the acid was required to neutralise the alkali. In the appendix this is titration procedure 3.
- Calculate the molarity of the sodium hydroxide and its concentration in g/dm³.
- moles = molarity x volume (in dm³ = cm³/1000)
- moles HCl = 0.200 x (15.0/1000) = 0.003 mol
- moles HCl = moles NaOH (1 : 1 in equation)
- so there is 0.003 mol NaOH in 25.0 cm³
- scaling up to 1000 cm³ (1 dm³), there are ...
- 0.003 x (1000/25.0) = 0.12 mol NaOH in 1 dm³
- molarity of NaOH is 0.120 mol dm^-3 (or 0.12M)
- since mass = moles x formula mass
- and M_r(NaOH) = 23 + 16 + 1 = 40
- concentration in g/dm³ = molarity x formula mass
- concentration in g/dm³ is 0.12 x 40 = 4.80 g/dm³

Titration Calculation Example 12.4 analysing sodium hydroxide with standard sulfuric acid.
- Given the equation: 2NaOH_(aq) + H₂SO_4(aq) ==> Na₂SO₄ + 2H₂O_(l)
  - This is read as 2 mol + 1 mol reactants ===> 1 mol + 2 mol products
- 25.00 cm³ portions of a sodium hydroxide were titrated with a standardised solution of 0.75 mol/dm³ sulphuric acid solution using phenolphthalein indicator. In the appendix this is titration procedure 1.
- If the average titration was 17.70 cm³ of sulfuric acid, what is the molar concentration of the sodium hydroxide?
  - moles H₂SO₄ in titration = molarity H₂SO₄ x volume in dm³
  - moles H₂SO₄ = 0.75 x (17.70/1000) = 0.013275 mol
  - From the balanced equation, for every mole of H₂SO₄, two moles of NaOH react
  - Therefore moles NaOH = 2 x moles H₂SO₄
  - moles NaOH = 0.013275 x 2 = 0.02655 mol
  - molarity of NaOH = moles NaOH / volume in dm³
  - molarity NaOH = 0.02655 / (25.00/1000) = 1.062 mol/dm³

Titration calculation Example 12.2 analysing sulfuric acid with standard potassium hydroxide
- Given the equation: 2KOH_(aq) + H₂SO_4(aq) ==> K₂SO₄ + 2H₂O_(l)
  - This is read as 2 mol + 1 mol reactants ===> 1 mol + 2 mol products
- 20.0 cm³ of a sulphuric acid solution was titrated with a standardised solution of 0.0500 mol/dm³ (0.05M) potassium hydroxide.
- Using phenolphthalein indicator for the titration, the acid required 36.0 cm³ of the alkali KOH for neutralisation what was the concentration of the acid? In the appendix this is titration procedure 2.
  - moles = molarity x volume (in dm³ = cm³/1000)
  - mol KOH = 0.0500 x (36.0/1000) = 0.0018 mol
  - mol H₂SO₄ = mol KOH / 2 (because of 2 : 1 ratio in equation above)
  - mol H₂SO₄ = 0.0018/2 = 0.0009 (in 20.0 cm³)
  - scaling up to 1000 cm³ of solution = 0.0009 x (1000/20.0) = 0.0450 mol
  - mol H₂SO₄ in 1 dm³ = 0.0450
  - so molarity of H₂SO₄ = 0.0450 mol dm^-3 (0.045M)
  - since mass = moles x formula mass
  - and M_r(H₂SO₄) = 2 + 32 + (4x16) = 98
  - concentration in g/dm³ is 0.045 x 98 = 4.41 g/dm³

Two more questions originally designed for advanced A-level chemistry students, but ok for GCSE level chemistry students

(From my original A-level acid-alkali titration questions SET 1 and SET 2)

Q22 If it took 20.55 cm³ of 0.100 M HCl to neutralise 25.0 cm³ of an NaOH solution, calculate the molarity of the alkali.

ANSWERS to Q22 SET 2. volumetric questions (non-redox) ok for gcse

Q23 A standardised solution of sodium hydroxide had a concentration of 0.1025 mol dm^-3 (0.1025M). If 25.0 cm³ of a sulfuric acid solution required 17.65 cm³ of the NaOH to neutralise it, calculate the molarity of the acid.

ANSWERS to Q23 SET 2. volumetric questions (non-redox) ok for gcse

Practise exam questions on acid-alkali titrations for Advanced A-level chemistry students

(From my original A-level acid-alkali titration questions SET 1 and SET 2)

After each question there is a link to the fully worked out answers.

Q1 A solution of sodium hydroxide contained 0.250 mol dm^-3.

Using phenolphthalein indicator, titration of 25.0 cm³ of this solution required 22.5 cm³ of a hydrochloric acid solution for complete neutralisation.

(a) write the equation for the titration reaction.

(b) what apparatus would you use to measure out (i) the sodium hydroxide solution? (ii) the hydrochloric acid solution?

(c) what would you rinse your apparatus out with before doing the titration ?

(d) what is the indicator colour change at the end-point?

(e) calculate the moles of sodium hydroxide neutralised.

(f) calculate the moles of hydrochloric acid neutralised.

(g) calculate the concentration of the hydrochloric acid in mol/dm³ (molarity).

ANSWERS to Q1 SET 1. volumetric questions (non-redox)

Q2 A solution made from pure barium hydroxide contained 2.74 g in exactly 100 cm³ of water.

Using phenolphthalein indicator, titration of 20.0 cm³ of this solution required 18.7 cm³ of a hydrochloric acid solution for complete neutralisation. [atomic masses: Ba = 137, O = 16, H = 1)

(a) write the equation for the titration reaction.

(b) calculate the molarity of the barium hydroxide solution.

(c) calculate the moles of barium hydroxide neutralised.

(d) calculate the moles of hydrochloric acid neutralised.

(e) calculate the molarity of the hydrochloric acid.

ANSWERS to Q2 SET 1. questions

Q3 4.90g of pure sulphuric acid was dissolved in water, the resulting total volume was 200 cm³.

20.7 cm³ of this solution was found on titration, to completely neutralise 10.0 cm³ of a sodium hydroxide solution. [atomic masses: S = 32, O = 16, H = 1)

(a) write the equation for the titration reaction.

(b) calculate the molarity of the sulphuric acid solution.

(c) calculate the moles of sulphuric acid neutralised.

(d) calculate the moles of sodium hydroxide neutralised.

(e) calculate the concentration of the sodium hydroxide in mol dm^-3 (molarity).

ANSWERS to Q3 SET 1. questions

Q7 A 50.0 cm³ sample of sulphuric acid was diluted to 1.00 dm³.

A sample of the diluted sulphuric acid was analysed by titrating with aqueous sodium hydroxide.

In the titration, 25.0 cm³ of 1.00 mol dm^-3 aqueous sodium hydroxide required 20.0 cm³ of the diluted sulphuric acid for neutralisation.

(a) give the equation for the full neutralisation of sulphuric acid by sodium hydroxide.

(b) calculate how many moles of sodium hydroxide were used in the titration?

(c) calculate the concentration of the diluted acid.

(d) calculate the concentration of the original concentrated sulphuric acid solution.

ANSWERS to 7 SET 1. questions

Titration index * Full chemistry calculations index

Quiz 12 The basics of acid-alkali titration calculations Good exam practice questions for GCSE level

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