ACIDS

BASES

WATER AS A BASE

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

WATER AS AN ACID

H₂O + NH₃ → OH^- + NH₄⁺

Water is donating a proton to ammonia

PROTONS

CALCULATING THE pH OF STRONG ACIDS

Find concentration c = n / v

 - log [H⁺] = pH

 CALCULATE CONCENTRATION FROM pH

DILUTING STRONG ACIDS

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⁺] 

CALCULATING pH FOR STRONG ACIDS NEUTRALISED BY A STRONG BASE

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⁺]

K_w

K_w = [H⁺] [OH^-] / [H₂O] 

H₂O is omitted because it's very small.

This is the equilibrium constant for water 

1 x 10^-14

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

CALCULATING THE pH OF STRONG ACIDS DILUTED WITH WATER

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⁺]

CALCULATE pH OF WEAK ACIDS

Rearrange the equation for K_a to make [H⁺][A^-] = K_a [HA] 

[H⁺] = [A^-] So K_a = [H⁺]²  =  [H⁺][A^-] = [H⁺]²

Calculate [H⁺]²  then √[H⁺]² to find the answer for [H⁺] 

Calculate pH -log[H⁺] 

IN A STRONG ACID

WHEN DOING A CALCULATION IN REVERSE

WEAK ACIDS AND BASES

ENDOTHERMIC PROCESSES

EXOTHERMIC

CONJUGATE BASE

A substance that becomes an acid by gaining a proton

e.g NH₃ + HBr ↔ NH₄⁺ + Br^-

NH₄⁺ is the conjugate base

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

WHY IS WATER ALWAYS NEUTRAL? 

ESSENTIAL FEATURE OF ACID BASE REACTIONS

DEFINITION OF K_w

DEFINITION OF pH

H₃O⁺

H⁺ IONS

DIFFERENCE IN pH

WHY IS HCl A STRONG ACID? 

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

pK_a

 = -log K_a 

This is the antilog of the acid dissociation constant :

K_a = [H⁺][A^-] / [HA]

UNITS FOR K_a

moldm^-3 moldm^-3 /  moldm^-3

=  moldm^-3

VALUES OF pK_a

TITRATION CURVE - STRONG ACID STRONG BASE*

TITRATION CURVE - STRONG ACID WEAK BASE*

TITRATION CURVE - WEAK ACID WEAK BASE*

TITRATION CURVE - WEAK ACID STRONG BASE*

EQUIVILANCE POINT

TITRATION CURVES

HOW TO CALCULATE CONCENTRATION USING THE EQUIVILANCE POINT

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

INDICATORS

END POINT

SUITABLE INDICATORS

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 

COMMON INDICATORS

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

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

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

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

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

HALF NEUTRALISATION POINT

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

BUFFER SOLOUTIONS

ACIDIC BUFFERS

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 

HA_(aq) ↔ H⁺_(aq) + A^-_(aq)

[H⁺_(aq) ] = [A^-_(aq)]

It's a weak acid, so both are very small because most of the HA is undissociated

ADDING ALKALI

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

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

ADDING ALKALI

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^-. 

FUNCTION OF WEAK ACID IN A BUFFER

FUNCTION OF SALT IN BUFFER

CHANGE IN pH

MIXTURE OF WEAK ACID

At half neutralisation where pH = pK_a 

neutralise half the acid to get this buffer

BASIC BUFFERS*

EXAMPLE OF A BUFFER

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

CALCULATION OF BUFFERS

HA_(aq) ↔ H⁺_(aq) + A^-_(aq)

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

to calculate pH of buffers 

EXAMPLE OF USING 

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

Buffer contains 0.100moldm^-3 ethanoic acid and 0.100moldm^-3 sodium 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^-5 x 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

IN A BUFFER SOLUTION

CALCULATING pH OF A WEAK ACID AND STONG BASE

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