ACIDS AND BASES

Bronsted-Lowry acids

proton donors

Bronsted lowry acids release H⁺ ions when mixed in water

which combine with H₂O to form hydroxonium ions H₃O⁺

Bronsted Lowry base

proton acceptors

acids and base dissociate in water

• acids break up into positive charged ions

• bases break up into negative charged ions

the strength of the acid/ base depends on

their ability to dissociate

the stronger the acid/base

the more they dissociate/ ionise in water

weak acids like carboxylic/citric acids

only very slightly dissociate in water so only number of H+ form

weak acid equilibrium

lies heavily on the left

weak bases like ammonia

only slightly dissociate in water

weak base equilibrium

lies heavily on the left

acids can only get rid of protons

if a base is available to accept them

when water is added to acid

water acts as a base

water dissociates into hydroxonium ions AND hydroxide ions

so an equilibrium exists in water

you can write the equilibrium constant expression for the equilibrium of water

water dissociates in TINY amounts

so thr equilibrium lies HEAVILY on the left

waters concentration is so much larger that the cocnentration of H+ and OH-

that the concentration of water is considered a constant value

ionic product of water K_w

multiplying the Kc expression by the concentration of water produces ANOTHER constant

Kw value is the same for an aqueous solution at a certain temperature

FOR EXAMPLE: 298K → 1.00×10^-14 mol² dm^-6

the value of Kw changes

as the temperature changes

in PURE water the ration of H+ to OH- is 1:1

so [H⁺]=[OH^-]

pH scale

the measure of the hydrogen ion concentration in a solution

you can calculate the concentration of a solution from the pH

[H^+]=10^-pH

monoprotic acids like HCl and HNO₃

each molecule of acid releases 1 proton when it dissociates

each mole of monoprotic acid produces 1 mole of hydrogen ions

meaning the H+ conc is the same as the acid conc

diprotic acids like H2SO4

each molecule of acid will release 2 protons when it dissociates

diprotic acids produce 2 moles of hydrogen ions for each mole of acid

meaning the H+ conc is twice the conc of the acid

STRONG ACID EXAMPLES THAT DONATE 1 MOLE OF OH- PER MOLE OF BASE

• sodium hydroxide

• potassium hydroxide

[base]=[OH-]

0.2 mol dm-3 of sodium hydroxide has 0.2 mol dm-3 [OH-]

if you know [OH- and the Kw value you can calculate the {H+]

• find the value of Kw and [OH-]

• rearrange the equation to make [H+] the subject

• using the [H+] find the pH using: -log₁₀[H+]=pH

weak acids and bases dissociate only slightly in solution

so the [H+] isn’t the same as the [acid conc]- making it trickier to find the pH

acid dissociation constant

Ka

for weak aqueous acid, HA, you get the equilibrium

HA⇄H+ + A-, where only a small amount of HA dissociates

[HA]»[H+]

[HA at the start]≂[HA at equilibrium]

when dealing with WEAK acids you can assume H+ ions come from the acid

[H+]≂[A-]

finding the pH weak acids

• write expression for Ka

• rearrange the equation and substitute values known to find [H+]

• take the SQUARE ROOT of the number to find [H+]

• substitute [H+] into -log10 [H+] to find the pH

finding the concentration of weak acids

• substitute the pH value into 10^-pH to calculate [H+]

• write expression for Ka

• rearrange the equation to give the conc of the acid

• substitute the value of Ka and [H+] into the equation and solve it

find the Ka of weak acids

if you know the concentration and the pH you can sue them both to find Ka of a weak acid

the value of Ka varies massively between acids

sometimes its easier to find the Ka value from its logarithmic constant pKa

pKa=-log₁₀Ka

Ka=10^-pKa

questions may ask to calculate the pH of a weak acid and give the concentration of acid and the pKa value

things to ensure your titration results are as accurate as possible:

• measure the neutralisation volume as precisely as you possibly can to the nearest 0.05cm3

• its a good idea to repeat the titration at least 3 times to get a mean titre value

• don’t use anomalous results, must be within ∓0.1cm3

a pH curve of the titration can be drawn if you find the pH using a pH probe/meter

the pH of the solution is the equivalence point/ midpoint

calculating concentration of monoprotic acids

• write balanced equation for the titration reaction

• decide what you know already and what you need to know, usually you’re given 2 volumes and a concentration and you’ll need to work out the other concentration

• for one reagent, you’ll know both the concentrations and volume so you can calculate the number of moles

• use the molar ratios in the balanced equation to find out how many moles of the other reagent reacted

• calculate the unknown concentration using conc=(Mol×1000)÷volume

diprotic acids release 2 protons in a solution like ETHANEDIOIC ACID (HOOC-COOH) the reaction happens in 2 stages because 2 protons are removed from the acid separately,

this means when you titrate ethanedioic acid with a really strong base you get a pH curve with 2 equivelance points

buffer

a solution that resists changes in pH when a small amount of acid/alkali is added

there are 2 types of buffers

• acidic buffer

• basic buffer

acid buffers pH is less that 7

they contain a mixture of a weak acid with one of its salts

acid buffers resist pH change when

either a base/acid is added to the solution

a mixture of ethanoic acid + sodium ethanoate is an example of a buffer solution

in the solution there will be loads of undissociated ethanoic acid molecules CH3COOH and ethanoate ions CH3COO- from the salt

altering the concentration of H+/OH- means the equilibrium is able to shift to counteract the change

resisting an acid

• the large amount of CH3COO- ensure the buffer can cope with the addition of acid

• more CH3COOH is produced as the equilibrium shifts left to reduce the concentration of H+ so the pH doesn’t change

resisting a base

• the large amount of CH3COOH ensure the buffer can cope with the addition of a base

• CH3COOH dissociates to form H+ IONS SHIFTING THE EQUILIBRIUM RIGHT

basic buffers have a pH greater than 7

they contain a mixture of weak bases with its salt and can resist the addition of acids/base

a mixture ammonia as a weak base and ammonia chloride as a salt of ammonia is an example of a basic buffer

Basic buffer resisting an acid

• the large amount of ammonia ensure the buffer can resist the addition of acid

• when H+ is added to a solution it can react with OH- to make water

• this shifts the equilibrium right to replace the OH- used up

• water reacts with ammonia to produce NH4+ and OH-

basic buffers resisting a base

• a large amount of ammonium chloride ensure the buffer resist the addition of a base

• the extra OH- reacts with the NH4+ to form ammonia and water

• the equilibrium shifts left to remove OH- from the solution

acidic and basic vuffers also resist pH change when diluted by water

water slightly dissociates so the extra H+ and OH- puh the equilibrium the same amount in both directions leaving it unchanged

alkaline conditions makes hair rougher

• shampoos have a pH around 5;5 to keep hair smooth and shiny

• buffers are in shampoo to maintain the pH while washing hair

biological washing powders contain buffers

they maintain the pH at the right level for enzymes to work best

its vital blood stays at a pH near 7.4

so our system contain buffers

if given the Ka and concentration of the weak acid and its salt you can calculate the pH of an acidic buffer

• write expression for Ka

• rearrange the equation to give an expression for [H+]

• substitute the value of Ka and the concentration of the acid and its salt into the equation

• solve the equation to find a value for [H+]

• substitute the [H+] into -log₁₀[H+]

other ways to make buffers

mixing a weak acid and a small amount of alkali to that some of the acid is neutralised is salt while some is left unneutralised

this reaction mixture would contain a weak acid and its salt to technically it is still an acidic buffer so you can work out its pH

• write out neutralisation equation

• calculate number of moles of acid/base at the start of reaction

• use the molar ratios in the equation to work out the moles of acid and salt left at the end of the reaction

• calculate the conc of the acid and salt in the buffer solution by dividing the moles at the end of the reaction by the volume of the solution

• then calculate the pH using Ka