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ACIDS

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An acid is a substance that can donate a proton (H⁺ ion)

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BASES

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A base is a substance that can accept protons

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WATER AS A BASE

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Can act as an acid and a base
base -
HCl + H₂O → H₃O⁺ + Cl^-
HCl donates a proton to the water, so the water acts like a base

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WATER AS AN ACID

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H₂O + NH₃→ OH^- + NH₄⁺

Water is donating a proton to ammonia

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PROTONS

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H⁺ ions - Hydrogen atom has only one electron, if this is lost all that remains in a proton. In water it's bonded to at least one water molecule to form the ion H₃O⁺

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CALCULATING THE pH OF STRONG ACIDS

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Find concentration c = n / v
;- log [H⁺] = pH

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;CALCULATE CONCENTRATION FROM pH

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SHIFT log(pH) = [H⁺];

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DILUTING STRONG ACIDS

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Calculate number of moles of acid n = c x v
Find new concentration by divinding moles with new volume c = n / v
Check if the acid is H₂SO_{4 ,} if so then multiply concentration by 2;
Calculate pH (-log[H⁺];

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CALCULATING pH FOR STRONG ACIDS NEUTRALISED BY A STRONG BASE

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Calculate moles of H⁺ ;n = c x v;
Calculate moles of OH^- n = c x v
Calculate moles in excess;nH⁺-;nOH^-;
Calculate concentration of [H⁺] c = n / v
pH = -log[H⁺]
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K_w

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K_w = [H⁺] [OH^-] / [H₂O];
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H₂O is omitted because it's very small.
This is the equilibrium constant for water
1 x 10^-14

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CALCULATE THE pH WHEN THE CONCENTRATION OF OH^- IS GIVEN

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[H⁺] = K_{w (1 x 10^-14)} / [OH^-]

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CALCULATING THE pH OF STRONG ACIDS DILUTED WITH WATER

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Calculate moles of OH^- n = c x v
Calculate new concentration of [OH^-] c = n / v
Calculate [H⁺] using K_w: [H⁺] = K_w / [OH^-]
Calculate pH: -log[H⁺]

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CALCULATE pH OF WEAK ACIDS

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Rearrange the equation for K_ato make [H⁺][A^-] = K_a [HA]
[H⁺] = [A^-] So K_a= [H⁺]²= [H⁺][A^-] = [H⁺]²
Calculate [H⁺]²then √[H⁺]²to find the answer for [H⁺]

Calculate pH -log[H⁺]

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IN A STRONG ACID

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The concentration of [H⁺] is equal to the concentration of acid

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WHEN DOING A CALCULATION IN REVERSE

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First thing you do when reversed is always the last thing you do

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WEAK ACIDS AND BASES

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Partly dissociate in aqueous solutions

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ENDOTHERMIC PROCESSES

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There will be a higher [H⁺] if the temperature increases so pH decreases

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EXOTHERMIC

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Lower concentration of H⁺ when temperature increases, so the pH is higher

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CONJUGATE BASE

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A substance that becomes an acid by gaining a proton
e.g NH₃ + HBr ↔ NH₄⁺ + Br^-
NH₄⁺ is the conjugate base

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K_w CONCENTRATION OF WATER: WHY IS H₂O OMITTED FROM EQUATION?

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The concentration of water is effectivel constant, becasue it very large so H₂O can be cancelled out

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WHY IS WATER ALWAYS NEUTRAL?

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Becase the [H⁺] = [OH^-] So they have the same number of moles, hence the same concentration

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ESSENTIAL FEATURE OF ACID BASE REACTIONS

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The transfer of protons

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DEFINITION OF K_w

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The ionic product of water

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DEFINITION OF pH

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The scale for measuring acidity and alkalinity. pH = -log [H⁺]

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H₃O⁺

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H⁺ions combine with water molecules to make the oxonium/hydronium/hydroxium ion

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H⁺IONS

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Have no electrons of thier own so it can only form a bond with another species that has a lone pair of elecetrons

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DIFFERENCE IN pH

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Difference in one pH number is equal to a tenfold difference in [H⁺]

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WHY IS HCl A STRONG ACID?

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In the gas phase it is covalent
In an aqueous solution it is totally ionic, so there are no molecules left so it's a strong acid

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pK_a

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;= -log K_a;
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This is the antilog of the acid dissociation constant :
K_a;= [H⁺][A^-] / [HA]

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UNITS FOR K_a

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moldm^-3;moldm^-3;/;^;moldm^-3

=;^;moldm^-3

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VALUES OF pK_a

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The smaller the value of;pK_a the stronger the acid

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TITRATION CURVE - STRONG ACID STRONG BASE*

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TITRATION CURVE - STRONG ACID WEAK BASE*

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TITRATION CURVE - WEAK ACID WEAK BASE*

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TITRATION CURVE - WEAK ACID STRONG BASE*

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EQUIVILANCE POINT

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Point at which sufficient base/acid has been added to neutralise the acid/base. The pH at equivilance point is not always exactly 7

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TITRATION CURVES

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Except from weak acid and weak base, there is a large and rapid change in pH at the equivilance point even though it may not be at pH 7

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HOW TO CALCULATE CONCENTRATION USING THE EQUIVILANCE POINT

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e.g we find the equivilance point of a titration is when 25cm³ of 0.0150moldm^-3 base, neutralised by 15.0cm³ acid. This shows that 25cm³ base has the same number of moles of;15.0cm³ acid.;

n = c x v:so we calculate the number of moles of base;
= 0.0150 x (25 x 10^-3) = 3.75 x 10^-4;(mol of acid and base)
c = n / v : so c_acid;= 3.75;x 10^-4 / (15 x 10^-3) = 0.025moldm^-3

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INDICATORS

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To find concentrations of a solution of acid or alkali - equivilance point has exactly the same number of moles of hydrogen ions and hydroxide ions

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END POINT

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The volume of alkali or acid added when the indicator just changes colour

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SUITABLE INDICATORS

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Sharp colour change(1 drop of acid/base) to give full colour change
End point of titration given by indicator ;= equivilance point;
Indicator gives distinct colour change;

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COMMON INDICATORS

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Universal - gradual colour change
Methyl orange - changes from red to yellow at pH 3- 5
Bromophenol blue - changes from yellow to blue at pH 4-5
Methyl red - changes from red to yellow at pH 4- 6
Bromothymol blue - changes from yellow to blue at pH 6-7
Phenolphthalein - changes from colourless to pink/red at pH 10 -11

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SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON STRONG ACID STRONG BASE*

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SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON WEAK ACID STONG BASE*

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SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON STRONG ACID WEAK BASE*

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SUITABLITY OF PHENOLPHTHALEIN AND METHYL ORANGE ON WEAK ACID WEAK BASE*

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HALF NEUTRALISATION POINT

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Point half way between zero and the equivilance point; pH will change very little when we add acid/base up to this point
Here we can find pK_a;of a weak acid because at this point
HA + OH^-;; H₂O + A^-;
At this point: [HA] = [A^-;]
;so,;K_a; = [H⁺] [A^-] / [HA];
K_a;= [H⁺];
So -logK_a;= -log [H⁺] So pK_a;= pH

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BUFFER SOLOUTIONS

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They resist changes of acidty and alkalinity when small amounts of acid or alkali are added to them, thier pH remains almost constant; they keep the concentration of hydrogen ions and hydroxide ions almost unchanged; based on equilibrium reaction which moves direction if either ion is added

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ACIDIC BUFFERS

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Made from weak acids because dissociation of weak acids is an equilibrium reaction
Definition: An acidic buffer is made from a mixture of a weak acid and a soluble salt of that acid. It will maintain a pH of below 7;

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HA_(aq);; H⁺_(aq) + A^-_(aq)

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[H⁺_(aq);] = [A^-_(aq)]
It's a weak acid, so both are very small because most of the HA is undissociated

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ADDING ALKALI

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HA_(aq) + OH^-_(aq)→ H₂O_(aq) + A^-_(aq)

The OH^- is removed so the pH tends to stay the same

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ADDING ALKALI

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Adding H⁺shifts the equilibrium to the left, so they combine with the A^-ions to make HA . However, since there is limited supplies of just A^-, once no more can be combined and there is an excess in H⁺ ions, the pH changes.
Soluble salts of HA can be added that ionise and increases the supply of A^-.

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FUNCTION OF WEAK ACID IN A BUFFER

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Acts like HA which can remove OH^-

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FUNCTION OF SALT IN BUFFER

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Source of A^- ions that can remove any added H⁺ ions

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CHANGE IN pH

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Buffers don't ensure NO change in pH at all. Only slight changes however. A buffer can be saturated where all avaliable HA or A^-;is used up

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MIXTURE OF WEAK ACID

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At half neutralisation where pH = pK_a;
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neutralise half the acid to get this buffer

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BASIC BUFFERS*

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Resist change, maintian pH above 7; made from a mixture of a weak base and the salt of that base

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EXAMPLE OF A BUFFER

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Blood approx 7.4 because even a small change such as 0.5 can be fatal. It's buffered by this equation
H⁺_(aq) + HCO^-_3(aq)↔ CO_2(aq)+ H₂O_(l)
adding extra H⁺ shifts equilibrium to right
Adding extra OH^-shifts equilibrium to left

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CALCULATION OF BUFFERS

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HA_(aq) ↔ H⁺_(aq)+ A^-_(aq)
K_a= [H⁺_(aq)] [A^-_(aq] / [HA_(aq)]
to calculate pH of buffers

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EXAMPLE OF USING
K_a= [H⁺_(aq)] [A^-_(aq] / [HA_(aq)]

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Buffer contains 0.100moldm^-3 ethanoic acid and 0.100moldm^-3sodium ethanoate. K_a = 1.7 x 10^-5 and pK_a = 4.77 for ethanoic acid.
K_a= [H⁺_(aq)] [A^-_(aq] / [HA_(aq)] Sodium ehtanote - fully dissociated , ethanoic acid, almost undissociated.
[H⁺] = K_a[HA] / [A^-] = (1.7 x 10^-5x 0.100) / (0.100) = 1.7 x 10^-5
-log(1.7 x 10^-5) = 4.77
pH = 4.77 pKa= 4.77
pH = pK_a with equal conc of acid and base

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IN A BUFFER SOLUTION

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[H⁺] DOESN'T EQUAL [A^-] SO SIMPLIFIED EXPRESSION K_a;= [H⁺]^2;/ [HA] IS NOT VALID

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CALCULATING pH OF A WEAK ACID AND STONG BASE

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Work out moles of both acid and base;
Find XS moles of base
Work out concentration from the moles with the new volume
Rearrange K_w for [H⁺] and calculate the concentration of hydrogen ions
Calculate pH
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