Acids salts and bases Flashcards
Calculate the concentration of solutions in g dm -3
● Concentration of a solution can be measured in mass per given volume of
solution e.g. grams per dm 3 (g/dm 3 )
● to calculate concentration of a solution use the equation
concentration (g dm -3 ) = mass of solute (g) volume (dm 3 )
● To calculate mass of solute in a given volume of a known concentration use the
equation: mass = conc x vol i.e. g = g/dm 3 x dm 3 (think about the units!)
number of moles = ____(mol/dm^3) * volume(___)
conc
dm^3
● Acids produce _____ in aqueous solutions
● Alkalis produce ______ in aqueous solutions
H + ions
OH - ions
Recall that a neutral solution has a pH of _ and that acidic solutions
have ____ pH values and alkaline solutions ______ pH values
7
lower
higher
Recall the effect of acids and alkalis on indicators, including litmus,
methyl orange and phenolphthalein
Phenolphthalein
○ Alkaline = pink
○ Acidic = colourless
● Methyl orange
○ Alkaline = yellow
○ Acidic = red
● Litmus paper
■ Blue litmus paper goes red in acidic & stays blue in alkaline
■ Red litmus paper goes blue in alkaline & stays red in acidic
Recall that as hydrogen ion concentration in a solution
increases by a factor of 10, the pH of the solution
decreases by 1
As the pH decreases by one unit, the H + concentration of the solution increases
by a factor of 10.
Investigate the change in pH on adding powdered
calcium hydroxide or calcium oxide to a fixed volume of dilute hydrochloric
acid
● method:
○ add dilute HCl to the beaker and measure pH
○ add weighed mass of calcium hydroxide and stir then record pH
○ keep adding weighed masses of calcium hydroxide until there is no more
change to the pH
● analysis:
○ draw a line graph with mass added on the horizontal axis and with pH on
the vertical axis
○ draw a line of best fit (remember to ignore any anomalies)
Explain the terms dilute and concentrated, with respect to
amount of substances in solution
▪ Strong and weak is NOT the same as concentrated and dilute – the latter refers
to the amount of substance whereas, the former refers the H + ion conc. in aq.
solutions
▪ Concentrated = larger amount of substance in a given volume of a solution
▪ Dilute = lesser amount of substance in a given volume of a solution
Explain the terms weak and strong acids
▪ Strong acid = fully dissociates in aqueous solution (dissociation is where an acid
breaks down to release H + ions in solution)
o e.g. hydrochloric, nitric and sulfuric acids
▪ Weak acid = partially dissociates in aqueous solution
o e.g. ethanoic, citric and carbonic acids
▪ Stronger an acid, greater the dissociation, the more H + ions released, the lower
the pH (for a given conc. of aq. solutions)
Recall that a base is any substance that
reacts with an acid to form a salt and water only
acid + base → salt + water
Recall that alkalis are
soluble bases
acid + metal
acid + metal oxide
acid + metal hydroxide
acid + metal carbonate
→ salt + hydrogen gas (H 2 )
→ salt + water
→ salt + water
→ salt + water + carbon dioxide (CO 2 )
Describe the chemical test for: hydrogen and carbon dioxide (using
limewater)
● Test for hydrogen:
o Use a burning splint held at the open end of a test tube of the gas
▪ Creates a ‘squeaky pop’ sound
● Test for carbon dioxide:
o Bubble the gas through the limewater (calcium hydroxide solution) and it
will turn milky (cloudy)
Describe a neutralisation reaction as a reaction between
an acid and a base
● for any neutralisation reaction with an acid and an alkali the ionic equation is:
H + (aq) + OH - (aq) -> H 2 O(l)
Explain why, if soluble salts are prepared from an acid and an insoluble
reactant: excess of the reactant is added, the excess reactant is removed,
and the solution remaining is only salt and water
▪ Excess of the reactant is added
o this is to ensure your volume of acid reacts completely
▪ excess reactant is removed
o this is done by filtration of the insoluble reactant and is done so that you
are left with just a salt and water
▪ the remaining solution is only salt and water
o this is because all of your acid has fully reacted and you have filtered off
your other reactant, and that the only products of your reaction are a salt
and water
o if you have used a carbonate you would still only have a salt and water
remaining as carbon dioxide gas would have been given off into the
atmosphere
Explain why, if soluble salts are prepared from an acid and a soluble
reactant: titration must be used, the acid and the soluble reactant are then
mixed in the correct proportions, and the solution remaining, after reaction,
is only salt and water
● Titration must be used:
○ both reactants are liquids/soluble, so if you have an excess of one you
would not be able to easily remove it from your mixture of products, this
means you need to measure the exact amount of volumes that react,
which is easily done using a titration.
○ You can then mix the exact proportions of the two reactants
● The exact amount of acid has thus been added to the soluble reactant, meaning
that the leftover solution is only salt and water, no acid or alkali, because they
have been completely neutralised
Investigate the preparation of pure, dry hydrated copper sulfate crystals starting from copper oxide including the use of a water bath
● method:
○ add an excess of copper oxide (insoluble) to your acid (sulfuric acid-
H 2 SO 4 - as you are making copper SULFATE)
○ use a filter and filter paper to filter off any copper oxide that hasn’t
reacted (your solution should be blue as copper sulfate solution has been
formed)
○ evaporate off the water by placing your final solution in a water bath
Describe how to carry out an acid-alkali titration, using burette, pipette
and a suitable indicator, to prepare a pure, dry salt
How to carry out a titration:
1. Wash burette using the acid and then water
2. Fill burette to 100cm 3 with acid with the meniscus’ base on the 100cm 3 line
3. Use 25cm 3 pipette to add 25cm 3 of alkali into a conical flask, drawing alkali
into the pipette using a pipette filler
4. Add a few drops of a suitable indicator to the conical flask (eg:
phenolphthalein which is pink when alkaline and colourless when acidic)
5. Add acid from burette to alkali until end-point is reached (as shown by
indicator.
6. The titre (volume of alkali needed to exactly neutralise the acid) is the
difference between the first (100cm 3 ) and second readings on the burette)
7. 8. Repeat the experiment to gain more precise results
To prepare a pure, dry salt – you warm the salt solution to evaporate the
water
9. Crystals form
Recall the general rules which describe the solubility of common types
of substances in water:
sodium
potassium
ammonium
nitrates
chlorides
sulfates
carbonates
hydroxides
type of salt : soluble : insoluble
: all
: all
: all
: allall except… : silver, leadall except… : lead, barium, calciumsodium, potassium, ammonium : …all exceptsodium, potassium, ammonium : …all except
Predict, using solubility rules, whether or not a precipitate will be
formed when named solutions are mixed together, naming the precipitate if
any
● first, work out what the products of your reaction will be
● then, use the solubility rules to determine if any salts formed are soluble/insoluble
● any INSOLUBLE salts will form as a precipitate (as any soluble salts will remain in
solution)
Describe the method used to prepare a pure, dry sample of an insoluble
salt
mix the two solutions needed to form the salt
filter the mixture using filter paper, which the insoluble salt will be left on
wash the salt using distilled water
leave the salt to dry on filter paper (water will evaporate, speed this process up
by drying it in an oven)
