4.1.1 Acids, alkalis and the pH scale

Acids supply hydrogen ions and alkalis supply hydroxide ions

1. An acid dissolved in water is a source of hydrogen ions, $H^{+}$.
2. An alkali dissolved in water is a source of hydroxide ions, ${OH}^{-}$.
3. Those ions are present because the acid or alkali has dissolved and separated into ions.
4. Hydrogen ions make a solution acidic, and hydroxide ions make it alkaline.
5. Hydrochloric, sulfuric and nitric acid are the three acids met most often.
6. Sodium hydroxide, potassium hydroxide and calcium hydroxide are common alkalis.

The pH scale runs from acidic through neutral to alkaline

1. The pH scale runs from $0$ to $14$ and shows how acidic or alkaline a solution is.
2. A neutral solution has a pH of exactly $7$.
3. An acidic solution has a pH below $7$, and the lower the value the more acidic it is.
4. An alkaline solution has a pH above $7$, and the higher the value the more alkaline it is.
5. So pH $2$ is more acidic than pH $5$, and pH $12$ is more alkaline than pH $9$.
6. Universal indicator gives an approximate pH from its colour, while a pH probe gives a number.
7. A pH probe is the more precise of the two, because it does not depend on judging a shade by eye.

Each indicator has its own set of colours

1. An indicator changes colour according to whether a solution is acidic, neutral or alkaline.
2. Litmus is red in acid, purple in neutral solution and blue in alkali.
3. Methyl orange is red in acid, orange in neutral solution and yellow in alkali.
4. Phenolphthalein is colourless in acid, colourless in neutral solution and pink in alkali.
5. Phenolphthalein therefore cannot tell an acidic solution from a neutral one.
6. A colour only means something once the indicator that produced it has been named.

Reading an indicator or a pH result

1. Name the indicator before reading anything into its colour.
2. Match that colour against the set of colours belonging to that indicator.
3. With a pH value, compare it with $7$: below is acidic and above is alkaline.
4. A pH of exactly $7$ is neutral, not almost acidic.
5. Two indicators pointing the same way is a check on the result rather than a contradiction.

4.1.2 Hydrogen ion concentration and pH

Hydrogen ion concentration sets the pH

1. pH measures the concentration of hydrogen ions, $H^{+}$, in a solution.
2. The scale is not a straight one: each step of $1$ in pH stands for a tenfold change in that concentration.
3. A higher $H^{+}$ concentration gives a lower pH.
4. A higher ${OH}^{-}$ concentration gives a higher pH.
5. A neutral solution at pH $7$ holds the two ions in equal concentrations.

![A diagram showing the pH scale from 0 to 14 and its relationship to [H+

$$and[OH-$$

concentrations. As pH increases, [H+

$$decreasesbyfactorsoften(10^{0}to10^{-}14mol/d{m}^3)while[OH-$$

increases. At pH 7, both concentrations are equal at 10^-7 mol/dm^3.](https://assets.mathsgenie.co.uk/content-studio/notion-illustrations/3a9142830a338130a1efd2371fd738b3/1-8ecdc823d42cdf00-ph-scale-and-ion-concentrations-3a914283-genie.png)

Each step of one on the scale is a factor of ten

1. Multiplying the $H^{+}$ concentration by $10$ lowers the pH by $1$.
2. Dividing the $H^{+}$ concentration by $10$ raises the pH by $1$.
3. So a solution at pH $5$ moves to pH $4$ when its $H^{+}$ concentration is multiplied by $10$.
4. A change from pH $4$ to pH $3$ therefore means the concentration rose ten times.
5. A change from pH $3$ to pH $1$ means it rose $100$ times, because that is two steps.
6. The steps work the same way in reverse, so pH $2$ to pH $5$ is a thousandfold fall.

Hydroxide ions push the pH the other way

1. Adding alkali raises the hydroxide ion concentration in the solution.
2. A higher ${OH}^{-}$ concentration gives a higher pH.
3. The two concentrations move in opposite directions, because $H^{+}$ and ${OH}^{-}$ react together to form water.
4. Multiplying the ${OH}^{-}$ concentration by $10$ raises the pH by $1$, mirroring the acid case.
5. Diluting an alkali lowers its ${OH}^{-}$ concentration, so its pH falls back towards $7$.

Working out a pH change

1. Decide which ion the question changes, $H^{+}$ or ${OH}^{-}$.
2. Count how many factors of ten the concentration changes by.
3. Each factor of ten is one unit of pH.
4. Set the direction: more $H^{+}$ lowers the pH, more ${OH}^{-}$ raises it.
5. Give the new pH, or the factor, according to what the question asks for.

4.1.3 Dilute and concentrated, weak and strong acids

Concentration and strength describe different things

1. Concentration describes how much acid is dissolved in a given volume of solution.
2. Strength describes what proportion of the dissolved acid molecules split up into ions.
3. The two are independent, so an acid can be dilute and strong, or concentrated and weak.
4. Adding water changes the concentration but never the strength.
5. Strength is a property of the acid itself, fixed by which acid it is.

Dilute and concentrated describe how much acid is present

1. A dilute acid holds a relatively small amount of acid in a given volume of solution.
2. A concentrated acid holds a relatively large amount in the same volume.
3. Adding water increases the volume without changing the amount of acid, so the solution becomes more dilute.
4. Removing water reduces the volume, so the solution becomes more concentrated.
5. Diluting an acid raises its pH, because the $H^{+}$ concentration falls.

Strong and weak describe how far the acid dissociates

1. A strong acid dissociates completely, so every molecule splits into ions in solution.
2. Hydrochloric acid is strong: $HCl(aq)\to H^{+}(aq)+{Cl}^{-}(aq)$.
3. A weak acid dissociates only partially, so most of it stays as whole molecules.
4. Ethanoic acid is weak: ${CH}_{3}COOH(aq)\rightleftharpoons H^{+}(aq)+{CH}_3{COO}^{-}(aq)$.
5. The reversible arrow shows that ions and undissociated molecules are both present at once.
6. At the same concentration a strong acid gives a higher $H^{+}$ concentration than a weak one.
7. The strong acid therefore has the lower pH of the two at equal concentration.

Describing an acid needs both words

1. Dilute and strong: little acid present, but all of it dissociated.
2. Concentrated and strong: a lot of acid present, all of it dissociated.
3. Dilute and weak: little acid present, and only some of it dissociated.
4. Concentrated and weak: a lot of acid present, but only some of it dissociated.
5. Give the two descriptions separately, because they answer different questions.
6. A concentrated weak acid can hold a higher $H^{+}$ concentration than a dilute strong one.

4.1.4 Bases, alkalis and the reactions of acids

A base reacts with an acid to give a salt and water

1. A base reacts with an acid to form a salt and water, and nothing else.
2. Metal oxides and metal hydroxides are bases, because those are their only products with an acid.
3. A base that dissolves in water is also called an alkali.
4. Every alkali is therefore a base, but a base that will not dissolve is not an alkali.
5. Metal carbonates also react with acids, but they give carbon dioxide as well, so they fall outside that definition.
6. In every case the metal gives the first part of the salt's name and the acid gives the ending.

The acid fixes the ending of the salt's name

1. Hydrochloric acid makes chloride salts.
2. Sulfuric acid makes sulfate salts.
3. Nitric acid makes nitrate salts.
4. The metal supplies the first word, so magnesium with hydrochloric acid gives magnesium chloride.
5. A word equation names the substances without formulae, which is enough to get the salt right.
6. Copper oxide with sulfuric acid gives copper sulfate and water.

Acids with metals give a salt and hydrogen

1. A suitable metal reacts with a dilute acid to give a salt and hydrogen: $$acid+metal\to salt+hydrogen$$
2. Magnesium with hydrochloric acid: $$Mg(s)+2HCl(aq)\to {MgCl}_2(aq)+H_2(g)$$
3. Magnesium with sulfuric acid: $$Mg(s)+H_2{SO}_4(aq)\to {MgSO}_4(aq)+H_2(g)$$
4. The bubbles seen during the reaction are hydrogen leaving the solution.
5. This is not a neutralisation, because hydrogen is a product alongside the salt.

Acids with metal oxides and hydroxides give a salt and water

1. Both a metal oxide and a metal hydroxide give only a salt and water with an acid:

   $$acid+metal\; oxide\to salt+water$$ $$acid+metal\; hydroxide\to salt+water$$

2. Copper oxide with sulfuric acid: $CuO(s)+H_2{SO}_4(aq)\to {CuSO}_4(aq)+H_{2}O(l)$.

3. Sodium hydroxide with hydrochloric acid: $NaOH(aq)+HCl(aq)\to NaCl(aq)+H_{2}O(l)$.

4. Calcium hydroxide needs two acid molecules: ${Ca(OH)}_2(aq)+2{HNO}_3(aq)\to {Ca(NO}_3{)}_2(aq)+2H_{2}O(l)$.

5. Both of these are neutralisations, because a salt and water are the only products.

Acids with metal carbonates also give carbon dioxide

1. A metal carbonate gives three products with an acid: $$acid+metal\; carbonate\to salt+water+carbon\; dioxide$$
2. Calcium carbonate with hydrochloric acid: ${CaCO}_3(s)+2HCl(aq)\to {CaCl}_2(aq)+H_{2}O(l)+{CO}_2(g)$.
3. Sodium carbonate with sulfuric acid: ${Na}_2{CO}_3(aq)+H_2{SO}_4(aq)\to {Na}_2{SO}_4(aq)+H_{2}O(l)+{CO}_2(g)$.
4. The fizzing is carbon dioxide escaping, and it is the quickest sign that a carbonate is present.
5. Hydrogen comes from a metal and carbon dioxide from a carbonate, so the gas identifies which one reacted.

4.1.5 Chemical tests for hydrogen and carbon dioxide

Two gases, two tests

1. A chemical test is a stated method together with the observation that counts as a positive result.
2. Hydrogen is tested with a burning splint.
3. Carbon dioxide is tested with limewater.
4. A test is only complete when both the method and the observation have been given.

Hydrogen: a burning splint gives a squeaky pop

1. Collect the gas in a test tube and keep it stoppered until the moment of the test.
2. Hold a burning splint at the open mouth of the tube.
3. Hydrogen burns rapidly in the air inside the tube.
4. That rapid burning makes a squeaky pop, and the sound is the positive result.
5. A splint that simply goes out has identified nothing, because no pop was heard.

Carbon dioxide: limewater turns milky

1. Limewater is a solution of calcium hydroxide, ${Ca(OH)}_2$.
2. Bubble the gas being tested through the limewater.
3. Carbon dioxide reacts with that calcium hydroxide to give insoluble calcium carbonate: $${Ca(OH)}_2(aq)+{CO}_2(g)\to {CaCO}_3(s)+H_{2}O(l)$$
4. The tiny solid particles make the limewater look milky, and that is the positive result.
5. Limewater that stays clear has not shown carbon dioxide.

Giving a complete test answer

1. Name the reagent or apparatus first.
2. Then give the observation that counts as a positive result.
3. Hydrogen: burning splint, squeaky pop.
4. Carbon dioxide: bubble through limewater, limewater turns milky.
5. Do not swap the two observations, because each belongs to one gas only.

4.1.6 Neutralisation as a reaction of acids with bases

Neutralisation is an acid reacting with a base

1. In a neutralisation an acid and a base react together to give a salt and water.
2. Both starting substances lose their acidic and basic character as the reaction proceeds.
3. An alkali is a soluble base, so an acid and an alkali neutralise each other in solution.
4. The general pattern is acid plus base giving salt plus water.
5. That pattern holds whichever acid and whichever base are used.

Hydrogen ions and hydroxide ions combine to make water

1. An acid in solution supplies hydrogen ions, $H^{+}(aq)$.
2. An alkali in solution supplies hydroxide ions, ${OH}^{-}(aq)$.
3. One of each holds exactly the atoms of one water molecule, so they react in a $1:1$ ratio: $$H^{+}(aq)+{OH}^{-}(aq)\to H_{2}O(l)$$
4. Removing those ions from the solution is what neutralisation means at the particle level.
5. The equation is the same for every acid and alkali, because each supplies one of the two ions.

The other ions are spectators

1. Sodium ions and chloride ions are present before the reaction and afterwards, unchanged.
2. Ions that do not change are left out of the ionic equation.
3. What remains is the change that actually happens, between $H^{+}$ and ${OH}^{-}$.
4. Those spectator ions stay in solution and become the salt once the water is evaporated.
5. The full equation shows the salt, while the ionic equation shows the change.

Writing about neutralisation

1. Identify the acid and say that it supplies $H^{+}$.
2. Identify the base or alkali and say that it supplies ${OH}^{-}$.
3. State that those two ions react together to form water.
4. Name the salt, which takes its name from the metal and from the acid.
5. Give the ionic equation with state symbols whenever the question asks for the particle-level change.

4.1.7 Preparing soluble salts

The method depends on whether the other reactant dissolves

1. A soluble salt stays in solution until the water is removed, so everything else has to be got rid of first.
2. When the other reactant is insoluble, it can be added in excess and filtered off afterwards.
3. When the other reactant is soluble, excess cannot be filtered off, so the volumes are matched by titration.
4. Choosing between the two routes starts with asking whether that reactant dissolves.
5. Either route finishes with crystallisation to obtain the dry salt.

Metals and carbonates reach the same salt by the same route

1. A metal added in excess gives the salt and hydrogen, and the leftover metal is filtered off.
2. A metal carbonate added in excess gives the salt, water and carbon dioxide, and the leftover solid is filtered off.
3. Zinc with sulfuric acid: $$Zn(s)+H_2{SO}_4(aq)\to {ZnSO}_4(aq)+H_2(g)$$
4. Copper carbonate with hydrochloric acid: $${CuCO}_3(s)+2HCl(aq)\to {CuCl}_2(aq)+H_{2}O(l)+{CO}_2(g)$$
5. Fizzing that stops is the sign that the carbonate route has finished.
6. Insoluble oxides, hydroxides, metals and carbonates can all be used this way.

Titration is used when both reactants dissolve

1. Rinse the pipette with the alkali and the burette with the acid, then discard both rinses.
2. Pipette a measured volume of the alkali into a conical flask.
3. Add a few drops of a single indicator such as methyl orange or phenolphthalein.
4. Fill the burette with acid, remove the funnel, and read the level at eye level to $0.05{cm}^3$.
5. Run acid in while swirling the flask, adding it dropwise as the colour begins to change.
6. Stop at the first permanent colour change and record the final reading.
7. The titre is the final reading minus the initial one.
8. Repeat until the titres agree within $0.10{cm}^3$, and average only those concordant results.

Getting a pure, dry salt from a titration

1. Repeat the titration with the mean volume of acid and no indicator in the flask.
2. The solution then holds only the salt and water, with no dye in it.
3. Transfer it to an evaporating basin and heat gently to concentrate it.
4. Leave the concentrated solution to cool so that crystals form.
5. Filter off the crystals and dry them between filter papers.
6. The salt is pure because the reactants were matched exactly and no indicator was present.

4.1.8 Solubility rules and preparing insoluble salts

The solubility rules, and the exceptions that matter

1. All common sodium, potassium and ammonium salts are soluble.
2. All nitrates are soluble.
3. Common chlorides are soluble, except silver chloride and lead chloride.
4. Common sulfates are soluble, except lead sulfate, barium sulfate and calcium sulfate.
5. Common carbonates are insoluble, except those of sodium, potassium and ammonium.
6. Common hydroxides are insoluble, except those of sodium, potassium and calcium, and calcium hydroxide is only slightly soluble.
7. The exceptions are the part worth learning, because the general rules follow a pattern.

Predicting whether a precipitate forms

1. Write down the ions present in each of the two solutions.
2. Pair each positive ion with the negative ion that came from the other solution.
3. Check both of those possible products against the solubility rules.
4. If one of them is insoluble it forms a precipitate, and it should be named.
5. If both are soluble, no precipitate forms and no reaction takes place.
6. Silver nitrate and sodium chloride: $${AgNO}_3(aq)+NaCl(aq)\to AgCl(s)+{NaNO}_3(aq)$$
7. Barium chloride and sodium sulfate: $${BaCl}_2(aq)+{Na}_2{SO}_4(aq)\to {BaSO}_4(s)+2NaCl(aq)$$
8. Potassium nitrate and sodium chloride give no reaction, because every possible product is soluble and all the ions simply stay in solution.

Ionic equations show only the ions that react

1. Only the ions that form the precipitate actually change.
2. The rest stay in solution throughout and are spectator ions.
3. For silver chloride the ionic equation is: $${Ag}^{+}(aq)+{Cl}^{-}(aq)\to AgCl(s)$$
4. For barium sulfate it is: $${Ba}^{2+}(aq)+{SO}_4^{2-}(aq)\to {BaSO}_4(s)$$
5. The state symbols carry the meaning here, because $(aq)$ and $(s)$ are what show the precipitation.

Preparing a pure, dry insoluble salt

1. Choose two soluble solutions that between them contain the ions you want.
2. Mix them and stir, so that the insoluble salt precipitates out.
3. Filter the mixture, keeping the residue on the filter paper.
4. Wash that residue with distilled water, which rinses away the soluble substances clinging to it.
5. Dry the washed solid between filter papers or in a warm oven.
6. The product is pure because everything else was soluble and has been washed away.