Acid-base neutralization and acid-base indicators.

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(topic12E) 2025-08-04

Titration, Acid-base neutralization
Contents
12.11.4.0 Titration, Acid-base neutralization
12.11.4.1 Acid-base neutralization, acid with base forms a salt and water
12.11.4.2 Acidity of vinegar and wine
12.11.4.3 Ammonia with sulfuric acid
12.11.4.5 Group analysis
12.11.4.10 Sodium hydroxide with hydrochloric acid
12.11.4.9 Simple titration of acids and bases
12.11.4.11 Titrate dilute hydrochloric acid with sodium hydroxide solution, with a burette
12.11.4.12 Titrate dilute sulfuric acid with sodium hydroxide solution and isolate sodium sulfate crystals
12.11.4.4 Carbon dioxide affects acid-base titration
12.11.4.6 Heat of neutralization titration
12.11.4.7 Microscale titration, sodium hydroxide with dilute acids
12.11.4.8 Prepare monoprotic acid solution from unknown molarity acid

12.11.4.1 Acid-base neutralization, acid with base forms a salt and water
In neutralization reactions, an acid and a base react in such proportions as to form a neutral solution of a salt and water.
In the home, wool dresses spotted with another colour from acids or bases can be restored to original colour by neutralization.
The reaction is between the hydrogen ions and the hydroxide ions.
H+ (aq) + OH- (aq) --> H2O.
Experiments
1. Add 2 mL nitric acid or tartaric acid to 4 cm water in a test-tube.
Add blue litmus paper, which then turns red.
Add drops of dilute ammonia solution or washing soda solution, Na2CO3.10H2O, but stop when the solution just turns blue again.
The acid is now neutralized by the alkali.
2. Put 3 mL dilute sodium hydroxide solution on a watch glass.
Use a dropper to add dilute hydrochloric acid drop by drop while stirring continuously.
Test the mixture with a fresh piece of litmus paper after each drop is added.
You can get a mixture where the litmus paper is neither red nor blue, but a tint midway between these two colours.
This mixture does not have the taste of an acid or the feel of an alkali.
The solution has the properties neither of an acid nor of an alkali.
The acid and alkali have neutralized each other.
Evaporate the solution to dryness by heating the watch glass over a beaker of boiling water.
A small quantity of solid appears on the watch glass.
This crystalline solid is sodium chloride, common salt.
Water is also a product of the reaction of sodium hydroxide with hydrochloric acid.
3. Pour about 5 mL of dilute solutions of sodium hydroxide, potassium hydroxide, calcium hydroxide, magnesium hydroxide and ammonia solution into test-tubes.
Neutralize each alkali solution with dilute hydrochloric acid then evaporate the resulting solutions to dryness.
Do not taste the residues.
Repeat the procedure with dilute sulfuric acid, dilute nitric acid or dilute acetic acid.
HCl (aq) + NaOH (s) --> NaCl (aq) + H2O (l)
HCl (aq) + KOH (s) --> KCl (aq) + H2O (l)
2HCl (aq) + Ca(OH)2 (s) --> CaCl2 (aq) + 2H2O (l)
2HCl (aq) + Mg(OH)2 (s) --> MgCl2 (aq) + 2H2O (l)
HCl (aq) + NH3 (aq) --> NH4Cl (aq) + H2O (l)
[ H2O (l) + NH3 (aq) --> NH4+ (aq) + OH- (aq)].
4. Add magnesium hydroxide in small amounts to dilute sulfuric acid until excess solid is present.
Filter the mixture and test the filtrate with pieces of red and of blue litmus paper.
Evaporate the filtrate to dryness.
Many metallic hydroxides react with acids to produce water and a salt in the same way as alkalis do.
magnesium hydroxide (s) + sulfuric acid (aq) --> water (l) + magnesium sulfate (aq)
Metallic hydroxides that behave in this way with acids and are insoluble in water are called basic hydroxides.
The three classes of compounds (alkalis, basic oxides and basic hydroxides) are representatives of a group of substances called bases.
H2SO4 (aq) + Mg(OH)2 (s) --> MgSO4 (aq) + 2H2O (l).

12.11.4.2 Acidity of vinegar and wine
Vinegar must have a minimum of 5% acidity if it is sold.
The percent acidity, % acidity (percent acetic acid) = (grams of acetic acid, CH3COOH / grams of vinegar) x 100
CH3COOH (aq) + NaOH (aq)--> CH3COONa (aq) + H2O (l)
Titrate 5.00 mL of household vinegar with sodium hydroxide solution.
The molarity of the sodium hydroxide solution is stated on the bottle label.
The molar mass of acetic acid (ethanoic acid) = 60.05.
The density of vinegar = 1.00 g / mL.
One mole of acetic acid reacts with one mole of sodium hydroxide
So the number of moles of acetic acid = molarity of the sodium hydroxide solution, in moles / litre X volume of sodium hydroxide solution, in litres.
Pour 5 mL of vinegar clean 250 mL Erlenmeyer flask and add 20 mL of distilled water and four drops of phenolphthalein indicator solution.
Add sodium hydroxide solution from the burette while swirling the flask, until the solution is a faint permanent pink and record the volume.
Discard the solution if you overshoot the end point and the solution is red.
Moles of acetic acid = (molarity of the sodium hydroxide solution, in moles / litre) x (volume of sodium hydroxide solution, in litres)
Grams of acetic acid = (moles of acetic acid) x (molar mass)
Grams of vinegar = (volume of vinegar) x (density of vinegar)
% acetic acid = (grams of acetic acid / grams of vinegar) x 100.

Measure acidity using a titration kit
To measure T.A. in wine, use an inexpensive titration or acid test kit.
Test kits can be purchased cheaply and can be used over and over again.
The amount of acid in wine is measured by slowly adding a small amount of the reagent base NaOH until a change in colour occurs from indicator phenolphthalein.
Put a 15 cc sample (one cc equals one ml) of wine into a test-tube.
Most test tubes that come with the acid test kits are marked with a line indicating this volume, or
use a small plastic syringe in the test kit to measure the desired amount into the test tube.
Then rinse the syringe.
Put 3 drops of the phenolphthalein solution into the test-tube.
Swirl or shake the test tube to mix the indicator with the wine.
Use the syringe to draw out 10 cc of the sodium hydroxide reagent, with no bubbles in the liquid.
Be careful! Avoid contact of the sodium hydroxide solution with your skin or eyes.
Add the sodium hydroxide solution to the test-tube, by 0.5 cc at a time.
After each addition of 0.5 cc, swirl or shake the test-tube to mix the contents.
The colour of the liquid will momentarily change upon the addition of the reagent.
If testing white wines, the colour change will be pink.
If testing red wines, the colour change will be grey.
Keep swirling the test-tube until the colour subsides.
If the colour of the wine returns to the original colour, repeat adding 0.5 cc until the colour change is permanent.
So when the colour, pink or grey, does not go away, stop and record the amount of reagent used.
To determine the acidity of the wine, for each cc of reagent used, equals 0.1 % TA.
For example, if you used 6 cc of sodium hydroxide to react with the wine, the titrarable acidity is 0.6 %.
Discard the sample, because it is now toxic.
Do not add the sample back into your original wine!
Wash and dry the test equipment before storing it.

12.11.4.3 Ammonia with sulfuric acid
Pass ammonia through sulfuric acid.
The common fertilizer ammonium sulfate or sulfate of ammonia forms.
2NH3 (g) + H2SO4 (aq) --> (NH4)2SO4 (aq) + 2H2O (aq)
H2SO4 (aq) + 2NH4OH (aq) --> (NH4)2SO4 (aq) + 2H2O (l).

12.11.4.5 Group analysis
A saturated solution can remain in equilibrium with undissolved molecules of the solute.
Two equilibria exist: 1. an equilibrium between the undissolved solute and dissolved molecules, and
2. an equilibrium between dissolved molecules and ions formed by dissociation
(XY) <--> XY <--> X+ + Y-
Equilibrium 1.: (XY) <--> XY
Equilibrium 2.: XY <--> X+ + Y-
(XY) = undissolved molecules
XY = dissolved, but unionized molecules
X+ + Y- = ions
The tendency of the solid to pass into solution depends on its active mass, solution pressure.
If the temperature remains constant, the active mass remains constant.
This happens because, by the law of mass action, concentration dissolved molecules / concentration undissolved molecules = the constant, K.
The concentration of dissolved molecules is also a constant.
By the law of mass action, if [concentration X+] [concentration Y-] / [concentration dissolved molecules] = a constant, in a saturated solution
Then the product of the concentrations of the ions is a constant.
If large concentrations of different ions are brought together into the same solution,
then ions of X+ and Y- will precipitate out of solution as solid molecules until the concentrations of the remaining ions in solution have the product of their concentrations equals the specific constant, the solubility product.
Group I
Lead, silver and mercury (I) are precipitated as chlorides by chloride ions from hydrochloric acid.
The concentrations of silver and chloride ions that can remain in solution are small.
When a solution of a silver salt containing silver ions is mixed with hydrochloric acid, most of the silver and chloride ions form molecular silver chloride and leave the solution as a solid phase until the remaining ions attain equilibrium.
(concentration silver ions) × (concentration chloride ions) = 1 × 10-10, the solubility product.
The solubility product has a constant value, so adding excess chloride ions reduces the concentration of the silver ions to a negligible quantity.
Only silver chloride, lead chloride and mercury (I) chloride have low solubility products.
So the ions of other metals remain in solution in the presence of high concentrations of chloride ions.
Group II
Assume hydrogen sulfide is ionized
H2S <--> 2H+ + S2-
By the law of mass action (concentration H+)2 × (concentration S2-) / (concentration unionized H2S) = the constant, 1.1 × 10-22.
In a neutral solution, the concentration of sulfide ions is low, because hydrogen sulfide is a weak electrolyte.
The concentration of hydrogen ions is also low.
In the acid solution used for Group II, the concentration of hydrogen ions is increased by the presence of the strong acid.
So to maintain the value of the solubility product constant, the concentration of the sulfide ion is reduced below its already small value in neutral solution.
However, the amount of sulfide ions is enough to allow the solubility products of the sulfides of mercury (II), lead, copper and bismuth to be exceeded.
Also, cadmium sulfide may precipitate if the acid is not too concentrated.
So in Group II, all the sulfides of mercury (II), lead, copper, bismuth and cadmium precipitate.
Solubility products:
Lead sulfide 4 × 10-28, Copper sulfide 8 × 10-45, Mercury (II) sulfide 4 × 10-54,
Cadmium sulfide 3.6 × 10-29, Manganese sulfide 1.4 × 10-15, Zinc sulfide 1.2 × 10-24
The concentration of sulfide ion in acid solution is not enough to allow the higher solubility products of the sulfides of manganese, zinc, cobalt or nickel to be reached with any possible concentration of the metal ion, so these sulfides do not precipitate.
They precipitate later in Group IV, where the precipitating agent is the highly ionized salt, ammonium sulfide, and the concentration of the sulfide ion from it is high.
Group III
The precipitating agent is ammonia solution.
NH4OH <--> NH4+ + OH-
By the law of mass action: (concentration NH4+) × (concentration OH-) / (concentration unionized NH4OH) = a constant.
Ammonia solution is a weak base, so does not ionize much and the value of the constant is only about 2 × 10-5.
Most of the ammonia solution will be dissolved, but not ionized.
The small hydroxyl ion concentration in a solution of ammonia solution that is also fairly concentrated with respect to ammonium chloride is still large enough to cause a precipitation of the hydroxides of ferric iron, chromium and aluminium, but not great enough to precipitate the hydroxides of zinc, manganese, cobalt and nickel.
Manganese hydroxide may precipitate slightly if the concentration of ammonium chloride is not sufficiently great.
Group IV
In Group II the presence of hydrogen ions from the added acid reduces the concentration of sulfide ions, but this reduced value was enough to allow the solubility products of the metallic sulfides in the group to be exceeded.
In Group IV, hydrogen sulfide is added to a solution made alkaline with ammonia solution and so contains excess of hydroxyl ions.
By the law of mass action: (concentration H+)2 × (concentration S2-) / (concentration unionized H2S) = a constant.
The ionic product of water [H+] × [OH-1] = (10-14).
The hydroxyl ions from the ammonia solution lower the concentration of hydrogen ions causing an increased concentration of S2- ion.
In Group IV, the metal sulfides not already precipitated in Group II, because of their high solubility products, precipitate.
So the ionic concentration of the sulfide ion is controlled by variation of the concentration of the ions it is associated with, i.e. hydrogen ions.
Group V
The metals still remaining in solution include barium, strontium, calcium, magnesium, sodium and potassium.
Barium, strontium and calcium are precipitated as carbonates by the addition of CO32- ions in alkaline solution, because of the low values of the solubility products, [X2+] × [CO32-] = K< between 10-8 and 10-9.
The solubility product of magnesium carbonate is low, 10-5, and in neutral solution would be precipitated, but its precipitation is prevented in this group by the ammonium ions, mainly from
the ammonium chloride added before Group III.
With the large NH4+ ion concentration from this source, the concentration of CO32- ions is reduced below the concentration to precipitate magnesium as a carbonate.
[NH4+]2 [CO32-] / [NH4)2CO3] = K.

12.11.4.10 Sodium hydroxide with hydrochloric acid
Put 10 drops of dilute sodium hydroxide solution on a watch glass.
Add drops of dilute hydrochloric acid and stir.
Test the mixture with litmus paper after adding each drop of hydrochloric acid.
When the litmus is neither red nor blue, but between the two colours, stop adding drops of acid.
Wet the tip of the finger with the mixture.
Rub the mixture between the fingers.
It does not feel slippery, so the solution is not alkaline.
When the correct quantities of hydrochloric acid and sodium hydroxide are mixed a solution forms that has the properties neither of the acid nor of the alkali.
The acid and alkali have neutralized each other.
Evaporate the neutralized solution to dryness by heating the watch glass over a beaker of boiling water.
Crystals of sodium chloride appear on the watch glass.
HCl (aq) + NaOH (aq) NaCl (s) + H2O (l)
acid + alkali --> salt + water
Add dilute hydrochloric acid to dilute solutions of: sodium hydroxide, potassium hydroxide, calcium hydroxide aqueous ammonia solution.
Evaporate to dryness.
Describe the salt formed.
Repeat the experiment with: dilute sulfuric acid, dilute nitric acid, dilute ethanoic acid (acetic acid).

12.11.4.9 Simple titration of acids and bases
Titration is an experimental method for measuring the concentration of a solution.
Measure the volume of the solution "A" needed to react with a given volume of solution "B".
For HCl and NaOH titration, molarity "A" X volume "A" = molarity "B" X Volume "B".
The end point in a titration occurs when an indicator changes colour.
Use a medicine dropper or a teat pipette as a simple burette.
The drops must always be the same size.
Within experimental error, when the same dropper is used, the same number of drops of alkali is needed to neutralize the same number of drops of acid.
When the concentration of the acid is known, the concentration of the base can be estimated by comparing the numbers of drops of acid and drops of base that just react.
Drop 100 drops of water from a medicine dropper into a measuring cylinder.
Calculate the volume of one drop.
Measure 25 mL of 2 M sodium hydroxide solution in the measuring cylinder and pour into an evaporating dish.
Add 2 drops of phenolphthalein solution and note the red colour of the indicator.
Wash the medicine dropper with the 2 M hydrochloric acid to get rid of remaining sodium hydroxide.
Add 2 M hydrochloric acid a drop at a time to the solution in the evaporating dish.
Stir as each drop is added.
Note the number of drops added until the colour just disappears completely.
Calculate the volume of added acid.
Heat the solution until almost dry.
Use gentle heat to avoid spattering.
Describe the appearance of the residue.
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
40 + 36.5 --> 58.5 + 18
Weight of NaOH in 25 mL of 2M solution = (25 X 2 X 40 / 1000) g.
The weight of sodium chloride expected in the evaporating dish = 58.5 X (25 X 2 X 40 / 1000) / 40 = 2.925 g.

12.11.4.11 Titrate dilute hydrochloric acid with sodium hydroxide solution, with a burette
See diagram 12.8.4: Titration
1. Pour a hydrochloric acid solution of known concentration (such as 0.11mol / L) into a clear, dry burette until the liquid level is above the "0" line.
Fix the burette vertically with a burette clamp.
Rotate the stopcock carefully to set the lowest point of the liquid meniscus exactly to "0" and to make simultaneously the tapered portion of the burette full of the acid solution without any air bubbles in it.
Use a pipette to transfer 20 mL of the sodium hydroxide solution to a conical flask.
Add two drops of phenolphthalein to the flask.
The solution immediately turns red.
Stand the flask on a piece of white paper under the burette.
While adding drop by drop the acid solution from the burette, swirl the flask constantly so that mixing of the base and acid solutions is rapid and thorough.
Note any change in the solution colour.
The neutralization is exactly completed and the end point occurs when half a drop or one drop of the acid solution turns the pale red solution colourless in the flask immediately after swirling.
Stop the titration and record the burette meniscus reading.
Read the volume of the used hydrochloric acid solution.
Calculate the concentration of the sodium hydroxide solution according to the related chemical equation.
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l).

2. Use a burette containing 50 mL of 0.5 M sodium hydroxide.
Use a pipette to put 10 mL of 0.5 M hydrochloric acid in a beaker under the burette.
Add 2 drops of phenolphthalein to the beaker.
Stand the beaker on white paper under the burette containing sodium hydroxide.
Add a drop at a time of the sodium hydroxide from the burette and stir the beaker with a swirling motion.
Note the colour change when a drop of acid disappears after the solution is swirled.
The end point occurs when the drop does not change colour after swirling.
The solution is now neutral.
Test the neutral solution with litmus paper.
Pour 5 mL of the neutral solution into an evaporating dish.
Heat to dryness and weigh when cool.
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l).

12.11.4.12 Titrate dilute sulfuric acid with sodium hydroxide solution and isolate sodium sulfate crystals
Safety
* Wear protective gloves when handling corrosive chemicals.
* Handle acids and alkalis with care.
* If any alkali gets into your eyes or onto your skin, report to your teacher immediately, and wash the affected area under running water for at least three minutes.
* If any acid gets into your eyes, report to your teacher immediately, and flush your eyes with running water for at least three minutes.
* If any acid gets onto your skin, wash the affected area with plenty of water.
* Eye protection must be worn.
* Sodium hydroxide 2.0 M is corrosive.
* Sulfuric acid 1.0 M is an irritant.
1. Wash a burette as follows:
1.1 Use a filter funnel to fill a burette with 20 mL of water.
1.2 Hold the burette horizontally and rotate it slowly to wash the inner wall.
Open the stopcock to run out all the water into the sink.
Close the stopcock.
Repeat the above procedure with 1.0 M sulfuric acid.
2. Fill the burette as follows:
2.1 Fill the burette too below the zero mark with 1.0 M sulfuric acid.
2.2 Clamp the burette vertically in a stand.
2.3 Open the stopcock for a few seconds to fill the burette jet completely with the acid.
Do not leave any bubbles in the burette jet.
2.4 Take the initial burette reading (V1) = mL (to two decimal places).
3. Wash a pipette as follows:
3.1 Using a pipette filler, suck sufficient water into a pipette to fill part of the bulb.
(Never use your mouth to suck a pipette.)
3.2 Hold the pipette horizontally.
Rotate it slowly so that the water washes the inner wall, up to the graduation mark.
3.3 Allow the water to run out into the sink.
Repeat 3.1 to 3.3 using 2.0 M sodium hydroxide solution.
4. Titration:
4.1 Using the pipette filler and the washed pipette, transfer 25.0 cm of the sodium hydroxide solution into a clean conical flask.
4.2. Add 2 drops of methyl orange indicator to the conical flask.
Note the colour of the solution in the conical flask.
4.3 Run the 1.0 M sulfuric acid into the flask while swirling the flask continuously.
4.4 Continue adding the acid, until the solution in the flask just turns to an orange colour.
4.5 Take the final burette reading (V2) = mL (to two decimal places).
4.6 The volume of sulfuric acid needed to neutralize 25.0 mL of sodium hydroxide solution is (V2 - V1) mL.
5. Isolation of sodium sulfate crystals from solution:
5.1 Empty the conical flask and rinse it with water.
5.2 Use a pipette filler to pipette 25.0 mL of 2.0 M sodium hydroxide solution into the flask.
Do not add any indicator to allow preparation of pure sodium sulfate crystals.
5.3 From the burette, add the same volume of sulfuric acid as calculated from 4.6, i.e. (V2- V1) mL.
This results in complete neutralization to form a hot concentrated sodium sulfate solution and water.
5.4 Pour the resultant solution into a small beaker.
5.5 Boil the solution gently to concentrate it.
Heat with a small flame, until 1/3 of the solution is left.
Do not evaporate the solution until dry, because spitting will occur.
5.6 Every 10 seconds, dip a glass rod into the boiling solution and then take it out.
If the immersed end becomes "cloudy" within 5 or 6 seconds, stop heating.
The solution has now become concentrated enough to deposit crystals on cooling.
5.7 Leave the solution to cool overnight.
5.8 The next day, sodium sulfate crystals should have formed.
5.9 Filter the crystals from the remaining solution.
5.10 Wash the crystals with drops of distilled water from a plastic wash bottle.
5.11 Use a spatula to transfer the crystals onto a piece of filter paper or absorbent paper.
5.12 Dry the crystals by gently pressing them between filter paper of absorbent paper.
5.13 Store the crystals in a labelled enclosed container.

12.11.4.4 Carbon dioxide affects acid-base titration
Add 2 drops phenolphthalein to 100 mL of deionized water.
Add 2 drops of 0.1 M sodium hydroxide.
The reaction forms a red colour.
Swirl vigorously for one minute.
The red colour fades, because of absorption of carbon dioxide from the air.

12.11.4.6 Heat of neutralization titration
The end point occurs at maximum temperature.
Use 25 mL of dilute sodium hydroxide solution.
Note the original temperature.
Add 1 mL of 2 M hydrochloric acid, stir with a thermometer and note the temperature.
Continue to add 1 mL of the acid and note the temperature.
Use graph paper to plot temperature rise against volume of acid added.
Read from the graph the maximum temperature rise and volume of acid that neutralized the sodium hydroxide solution.
Calculation: 25 X 1 / 1000 X concentration of sodium hydroxide = volume of HCl X 1 / 1000 X 2.

12.11.4.7 Microscale titration, sodium hydroxide with dilute acids
See diagram 12.8.7: Microscale titration apparatus
Use a sodium hydroxide solution to titrate a standardized acid solution.
Use the sodium hydroxide solution to titrate an unknown acid.
Calculate the concentration of the unknown acid.
Conventional titration vs microscale titration:
Conventional titration requires burettes, bulb pipettes and litres of solutions.
Ten microscale titrations will use fewer solutions than used for one conventional titration.
Burettes are large and fragile so spillage and breakage of glassware occur sometimes.
When an operator using conventional titration does three measurements, a microscale operator can do six to ten measurements.
Microscale titration does require different hand and finger skills than conventional titration and involves some differences in the calculation methods.

12.11.4.8 Prepare monoprotic acid solution from unknown molarity acid
(A monoprotic acid donates one hydrogen ion (H+) per molecule when it dissociates in water.
A monoprotic acid, (HZ),dissociates according to the following equation:
HZ (aq) + H2O (l) ⇌ H3O+ (aq) + Z− (aq)
Monoprotic acids include: | Acetic acid | Formic acid | Hydrochloric acid | Nitric acid |
Diprotic acids include: | Sulfuric acid | Carbonic acid | Oxalic acid | Citric acid |)
1. Use two 2 mL graduated glass pipettes, graduated to 0.01 mL.
Attach disposable pipette tips.
If they fall off the ends of the pipettes, seal with silicone putty.
Attach a 5 mL plastic syringe to each pipette with silicone tubing.
Lubricate the syringe with glycerol.
Transfer 15 mL of acid solution into a wide neck bottle and one drop of phenolphthalein indicator.
Transfer 15 mL sodium hydroxide into another wide neck bottle.
Clamp the two pipettes on a stand with a double clamp.
Rinse then fill the pipettes by drawing solution up into the pipette with the syringe.
Record pipette volumes to 0.001 mL.
Use a 10 mL Erlenmeyer flask containing water for comparison when detecting the faint pink of the endpoint.
Use pipette volumes in excess of 1.000 mL to provide four significant figures in the volume measurements.
2. Pour 1.25 mL of 0.1 M monoprotic acid into a 10 mL Erlenmeyer flask.
Add 0.1 M sodium hydroxide until the colour changes to a faint pink.
Record the final pipette volumes.
Drop volume is less than 0.02 mL.
Pour 1.25 mL of the unknown acid into a 10 mL Erlenmeyer flask.
Add 0.1 M sodium hydroxide until the colour changes to a faint pink.
Record the final pipette volumes For each titration, calculate the ratio of acid volume to alkali volume to allow concordance between titrations.
Discard discordant values and calculate the mean for accepted values.
Ratio 1 = volume 0.1 M monoprotic acid / volume 0.1 M sodium hydroxide
Ratio 2 = volume unknown acid / volume 0.1 M sodium hydroxide
Concentration of unknown acid = (Ratio 1 / Ratio 2) X concentration 0.1M monoprotic acid, e.g. If Ratio 1 = 1.158, Ratio 2 = 1.079, molarity of unknown acid = (1.158 / 1.079) X 0.1 = 0.1073 M.
Addition of a drop of the indicator to the acid solutions being measured dilutes the acids, e.g. 0.02 mL of indicator solution added to 15 mL of acid solution dilutes the acid to affect the measured.
concentration at the fourth significant figure.
However, this titration is done twice, with the "known" standard acid and with the "unknown" acid, so the errors will cancel at the fourth significant figure when the acids do not differ greatly in concentration.