pH

Question 1

Why is a logarithmic scale used for the pH scale?

Answer 1

• aqueous solutions can have a H+(aq) ion concentration ranging from 10moldm-3 to 10-15 moldm-3

Logarithmic scale:

• allows the concentrations of H+(aq) ions to be conveniently expressed
• typically numbers with negative indices varying over a very wide range

Question 2

Logarithms to the base 10

Answer 2

log10 10^x = x

For Example:

log10 10^3 =3

Question 3

The pH scale

Answer 3

Low pH = high [H+(aq)]

High pH = low [H+(aq)]

A pH increase in 1 represents a 10 fold decrease in [H+(aq)]

Question 4

Define:

• pH
• [H+(aq)]

Answer 4

pH = -log10[H+(aq)]

[H+] = 10^-pH

Question 5

Calculating pH when solution is diluted

Answer 5

• If volume DOUBLES then concentration HALVES

E.g

• 50cm3 is diluted to 100cm3
• concentration halves from 0.1moldm3 to 0.05moldm3

Halving the concentration of a STRONG acid increases its pH by 0.30

Question 6

The STRENGTH of acids

Answer 6

The EXTENT to which it dissociates into H+(aq) and A-(aq) ions

STRONG Acids:

FULLY DISSCOIATE in aqueous solution

HA(aq) — H+(aq) + A-(aq)

• HCl
• HNO3
• H2SO4
• HBr

WEAK Acids:

PARTIALLY DISSOCIATE in aqueous solution

HA(aq) (reversible sign) H+(aq) + A-(aq)

• equilibrium lies to far left as acid partially dissociates
• concentration of dissociate HA is high

Question 7

pH & [H+] of STRONG ACIDS

Answer 7

pH = -log10[H+(aq)]

[H+(aq)] = 10^-pH

Question 8

pH & [H+] of WEAK ACIDS

Answer 8

[H+] = square root of (Ka x [HA])

pH = -log10[H+]

Question 9

WEAK ACID:

Why is concentration in mol dm-3 not used to determine pH?

Answer 9

Not ALL the acid dissociates

• Ka is used, where pKa relates to Ka in the same way pH does to [H+]

pKa = -log10Ka

Ka = 10^-pKa

Question 10

WEAK ACID:

Acid dissociation constant, Ka

Answer 10

Indicates the extent of acid dissociation

Ka = [products] / [reactants]

  = [H+][A-] / [HA]
  = mol dm-3

LARGE Ka:
- large amount of dissociation
- [product] is high compared to [reactant]
- STRONGER WEAK ACID

SMALL Ka:
- small amount of dissociation
- [reactants] is high compared with [products]
- WEAKER ACID

Question 11

WEAK ACID:

What is the Ka equation simplified to?

2 APPROXIMATIONS made

Answer 11

Ka = [H+]^2 equilibrium /[HA] start

1) [H+] = [A-] at equilibrium

2) [HA] at eqm = [HA] undissociated

Question 12

WEAK ACID:

First approximation & its validity

          (explained)

Answer 12

• Dissociation of HA gives a 1:1 equilibrium concentration of H+ & A- so top of fraction becomes [H+]^2
• Assumes the dissociation of water is negligible
• at 25 degrees [H+]=10^7 mol dm-3
• at pH>7, [H+] from the water becomes more significant than the [H+] arising from the dissociation of the weak acid

The approximation breaks down when the acid is either very strong or very dilute

Question 13

WEAK ACID:

Second approximation & its validity

        (explained)

Answer 13

• The decrease in the concentration of HA through its dissociation is negligible
• i.e [HA]eqm»_space;[H+]eqm
• Hence : [HA]eqm=[HA]start
• Holds for weak acids with SMALL Ka values
• Breaks down when the [H+] becomes larger and there is a real difference between [HA]start and [HA]start - [HA] eqm

INVALID FOR:

• Stronger weak acids ~ Ka > 10^-2 moldm-3
• Very dilute solutions ~ [H+] value tends towards zero in Ka

Question 14

The equilibrium in Water

Answer 14

• Water molecules can behave as both an ACID or a BASE

Hydronium ion, H3O+(aq) ~ very strong acid

Hydroxide ion, OH-(aq) ~ very strong base

H2O(l) + H2O(l) (reversible sign) H3O+(aq) + OH-(aq)

simplifies to:

H2O(l) (reversible sign) H+(aq) + OH-(aq)

Question 15

Kw

Answer 15

Kw = [H+(aq)][OH-(aq)]

units: mol2 dm-6

Question 16

pH of water

Answer 16

At 25 degrees, water has a pH of 7:
[H+(aq)] = 10-pH = 10-7 mol dm-3

The ionisation of water leads to a 1:1 molar ratio of H+ to OH- so:[OH-(aq)] = [H+(aq)]

Hence, 25 degrees:

[OH-] = 10-7 mol dm-3

Kw = [H+][OH-]
= 1x10-14 mol2 dm-6 at 25 degrees

Question 17

Define the ionic product of water, include its value and units at 25 degrees

Answer 17

Kw = [H+(aq)][OH-(aq)]

= 1.00 x 10-14 mol 2 dm-6 at 25 degrees

Question 18

Relationship between pKw & Kw

Answer 18

pKw = -log10Kw

What is the value of pKw at 25 degrees?

pKw = -log10Kw

= - log(1x10-14)
= 14

Question 19

The relationship between [H+] & [OH-]

Answer 19

Kw = [H+][OH-]
= [H+]^2
= 1.00x10-14 mol2 dm-6 at 25 degrees

Kw = [H+][OH-]
10-14 = 10-8 x [OH-]
[OH-] = 10-6 mol dm-3

Indices of [H+] & [OH-] adds up to -14 for integer pH value

Question 20

Working out pH of STRONG BASES

       TWO METHODS

Answer 20

Kw:

[H+] = Kw / [OH-]
pH = -log10[H+]

pOH:

pOH = -log10[OH-]
pH = 14 - pOH

Question 21

Working out pH of STRONG BASES

Method 1 ~ Kw

Answer 21

Kw = [H+][OH-]

[H+] = Kw/ [OH-]

= 1.00 x 10-14 / [OH-]

pH = -log [H+]

Question 22

Working out pH of STRONG BASES

Method 2 ~ pOH

Answer 22

pOH = -log10[OH-]

pH + pOH =14

pH = 14 - pOH

Question 23

Why are pKa values used instead of Ka values when comparing the strength of acids?

Answer 23

Easier to compare because they do not have:

• negative indices
• vary over larger magnitudes

(unlike Ka values)