Concentration of Solutions

Lessons

Senior 3 · Chemistry

Concentration of Solutions

5 lessons

Lesson 4 of 5

Titration of Sodium Hydroxide with Hydrochloric Acid

25 min

Two liquids that look like water

Dilute sodium hydroxide and dilute hydrochloric acid are both colourless. Mix them and nothing seems to happen, although the beaker gets slightly warm as they neutralise each other. How, then, can you tell the exact moment when just enough acid has been added? Add a few drops of an indicator, deliver the acid a drop at a time from a burette, and watch for one drop that changes the colour. This technique, a titration, can find the concentration of a solution to within about 1%, and water-testing, food and medicine laboratories use it every day.

Definition

Titration: a method of finding the concentration of a solution by measuring the exact volume of it that reacts with a known amount of another solution.

Titre: the volume delivered from the burette in one titration (final reading minus initial reading).

Indicator: a dye whose colour depends on whether the solution is acidic or alkaline. Example: phenolphthalein is pink in alkali and colourless in acid.

The apparatus

Rinse the pipette with a little of the sodium hydroxide, and the burette with a little of the hydrochloric acid, so that no water is left to dilute them.
Use the pipette and filler to put exactly 25.0 cm³ of sodium hydroxide into a clean conical flask. Add 2–3 drops of indicator.
Fill the burette with acid, with no air bubble in the jet, and read the initial reading at the bottom of the meniscus (Figure 4.2).
Run in the acid while swirling the flask. Near the end, add it drop by drop, until one drop makes the colour change permanently. Read the final reading.
The first, quick run is a rough titration. Repeat carefully until you have at least two titres that agree within 0.10 cm³ (concordant titres), and average those.

A conical flask at the end point of a titration with phenolphthalein. The solution is only faintly pink: one more drop of acid would make it colourless. Photo: Kengksn, CC BY-SA 4.0, via Wikimedia Commons

End point and equivalence point

The reaction is a neutralisation. Its ionic equation shows what really happens:

$\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)} \qquad \mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$

Table 4.1: Two common indicators

Indicator	Colour in alkali	Colour in acid	At the end point
phenolphthalein	pink	colourless	pink just disappears (acid added to alkali)
methyl orange	yellow	red	orange

For a strong acid with a strong alkali, such as hydrochloric acid with sodium hydroxide, either indicator works, because the colour changes within one drop of the exact neutral point.

Definition

Equivalence point: the point at which the amounts of acid and alkali added are exactly those in the balanced equation, so neither is in excess.

End point: the point at which the indicator changes colour, so you stop adding from the burette.

A well-chosen indicator makes the end point fall within one drop (about $0.05\,\mathrm{cm^3}$ ) of the equivalence point. Example: with phenolphthalein, the drop of acid that turns the faint pink colourless marks both points.

Results and calculation

$25.0\,\mathrm{cm^3}$ of sodium hydroxide solution of unknown concentration was titrated with $0.100\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid, using phenolphthalein.

Table 4.2: Titration results (illustrative)

Titration	rough	1	2	3
final reading (cm³)	24.60	24.30	24.45	24.15
initial reading (cm³)	0.00	0.10	0.35	0.00
titre (cm³)	24.60	24.20	24.10	24.15

Titres 1, 2 and 3 agree within 0.10 cm³, so they are concordant; the rough titre is left out. Mean titre $= \frac{24.20 + 24.10 + 24.15}{3} = 24.15\,\mathrm{cm^3}$ .

Worked example

Problem. Use the mean titre in Table 4.2 to find the concentration of the sodium hydroxide in $\mathrm{mol\,dm^{-3}}$ and in $\mathrm{g\,dm^{-3}}$ . (NaOH = 40)

Solution.

Step 1 , moles of the known solution: $n(\mathrm{HCl}) = 0.100 \times \frac{24.15}{1000} = 0.002415\,\mathrm{mol}$

Step 2 , mole ratio from the equation: NaOH : HCl = 1 : 1, so $n(\mathrm{NaOH}) = 0.002415\,\mathrm{mol}$

Step 3 , concentration of the unknown:

$c(\mathrm{NaOH}) = \frac{0.002415}{0.0250} = 0.0966\,\mathrm{mol\,dm^{-3}}$

Mass concentration $= 0.0966 \times 40 = 3.86\,\mathrm{g\,dm^{-3}}$ .

Worked example

Problem. $25.0\,\mathrm{cm^3}$ of $0.100\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide is exactly neutralised by $12.50\,\mathrm{cm^3}$ of dilute sulfuric acid. Find the concentration of the acid.

Solution. $\mathrm{2NaOH(aq) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)}$

$n(\mathrm{NaOH}) = 0.100 \times 0.0250 = 0.00250\,\mathrm{mol}$ . The ratio is 2 : 1, so $n(\mathrm{H_2SO_4}) = \frac{0.00250}{2} = 0.00125\,\mathrm{mol}$ .

$c(\mathrm{H_2SO_4}) = \frac{0.00125}{0.01250} = 0.100\,\mathrm{mol\,dm^{-3}}$

The acid needed only half the volume because each $\mathrm{H_2SO_4}$ gives two $\mathrm{H^+}$ ions.

Warning

Common mistakes and safety. (1) Including the rough titre in the mean. (2) Forgetting the mole ratio: $c_{1} V_{1} = c_{2} V_{2}$ works only when the ratio is 1 : 1. (3) Reading the top of the meniscus, or reading from above or below eye level. (4) Sodium hydroxide, even dilute, can damage the eyes: wear eye protection and always use a pipette filler, never your mouth. If alkali or acid gets on your skin or in your eyes, rinse with plenty of running water for at least 15 minutes and get help.

Tip

Three steps, every time: moles of the known ( $n = c V$ ), mole ratio from the balanced equation, then concentration of the unknown ( $c = \frac{n}{V}$ ). Write burette readings to two decimal places, ending in 0 or 5.

Royal Society of Chemistry: Practical skills, titration procedure. Watch how the burette is filled and read, how the flask is swirled, and how the acid is added drop by drop near the end point.

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Practice

Define (a) titre, (b) end point, (c) equivalence point.
Why is the burette rinsed with the acid it will hold, and not with distilled water?
A burette reads 1.25 cm³ at the start and 23.70 cm³ at the end. What is the titre?
A student's titres are 22.60 (rough), 22.35, 22.45 and 22.40 cm³. Which are concordant? Find the mean titre.
$20.0\,\mathrm{cm^3}$ of potassium hydroxide solution needs $16.00\,\mathrm{cm^3}$ of $0.150\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. Find the concentration of the potassium hydroxide.
$25.0\,\mathrm{cm^3}$ of sodium hydroxide is neutralised by $20.00\,\mathrm{cm^3}$ of $0.050\,\mathrm{mol\,dm^{-3}}$ sulfuric acid. Find the concentration of the sodium hydroxide.
Why is the flask stood on a white tile and swirled during the titration?
A student suggests using universal indicator for this titration. Evaluate the suggestion.

Show answersHide answers

(a) The volume delivered from the burette in one titration. (b) The point where the indicator changes colour. (c) The point where the acid and alkali have reacted in exactly the amounts shown by the equation.
Water left inside would dilute the acid and make the titre too large.
$23.70 - 1.25 = 22.45\,\mathrm{cm^3}$ .
22.35, 22.45 and 22.40 (all within 0.10 cm³). Mean $= \frac{22.35 + 22.45 + 22.40}{3} = 22.40\,\mathrm{cm^3}$ .
$n(\mathrm{HCl}) = 0.150 \times 0.01600 = 0.00240\,\mathrm{mol} = n(\mathrm{KOH})$ ; $c = \frac{0.00240}{0.0200} = 0.120\,\mathrm{mol\,dm^{-3}}$ .
$n(\mathrm{H_2SO_4}) = 0.050 \times 0.02000 = 0.0010\,\mathrm{mol}$ ; $n(\mathrm{NaOH}) = 2 \times 0.0010 = 0.0020\,\mathrm{mol}$ ; $c = \frac{0.0020}{0.0250} = 0.080\,\mathrm{mol\,dm^{-3}}$ .
The white tile makes the faint colour change easy to see; swirling mixes each drop of acid through the alkali, so the colour does not change too early in one spot.
A poor choice. Universal indicator changes colour gradually through many shades, so there is no single sharp colour change to mark the end point. Phenolphthalein or methyl orange gives a sharp change within one drop.

Sources and credits

Photo: Faint pink color of Phenolphthalein.jpg, Kengksn, licensed under CC BY-SA 4.0, via Wikimedia Commons.
Video: Practical skills assessment video, titration procedure, Royal Society of Chemistry, embedded from YouTube.
Written by Ganzaa. The diagrams, tables, worked examples and Practice items were drawn and written by Ganzaa; the results in Table 4.2 are illustrative.

1Two liquids that look like water

2The apparatus

3End point and equivalence point

4Results and calculation

Two liquids that look like water

The apparatus

End point and equivalence point

Results and calculation