pH

• The?pH?indicates the?acidity?or?basicity?of an acid or alkali
• The pH scale goes from 0 to 14
  • Acids have pH between 0-7
  • Pure water is?neutral?and has a pH of 7
  • Bases and alkalis have pH between 7-14
• The pH can be calculated using: pH = -log₁₀?[H⁺]

where [H⁺] = concentration of H⁺?ions (mol dm^-3)

• The pH can also be used to calculate the concentration of H⁺?ions in solution by rearranging the equation to:

[H⁺] = 10^-pH

Worked Example: Calculating the pH of acids

Answer

pH = -log [H⁺]

= -log 1.32 x 10^-3

= 2.9

K_a?& pK_a

• The?K_a?is the?acidic dissociation constant
  • It is the?equilibrium constant?for the dissociation of a?weak?acid?at 298 K
• For the?partial?ionisation?of a weak acid HA the equilibrium expression to find?K_a?is as follows:

HA (aq) ? H⁺?(aq) + A^-?(aq)

• When writing the equilibrium expression for weak acids, the following assumptions are made:
  • The concentration of hydrogen ions due to the ionisation of water is negligible
  • The dissociation of the weak acid is so small that the concentration of HA is approximately the same as the concentration of A^-
• The value of?K_a?indicates the extent of dissociation
  • A high value of?K_a?means that:
    • The equilibrium position lies to the right
    • The acid is?almost?completely?ionised
    • The acid is?strongly acidic
  • A low value of?K_a?means that:
    • The equilibrium position lies to the left
    • The acid is?only slightly ionised?(there are mainly HA and only a few H⁺?and A^-?ions)
    • The acid is?weakly acidic

• Since?K_a?values of many weak acids are?very low, pK_a?values are used instead to compare the strengths of weak acids with each other

pK_a?= -log₁₀?K_a

• The?less?positive?the pK_a?value the?more?acidic?the acid is

Worked Example: Calculating the?K_a?& pK_a?of weak acids

Answer

• Step 1:?Write down the equation for the partial dissociation of ethanoic acid

CH₃COOH (aq) ? H⁺?(aq) + CH₃COO^-?(aq)

• Step 2:?Write down the equilibrium expression to find?K_a

• Step 3:?Simplify the expression

The ratio of H⁺?to CH₃COO^-?is 1:1

The concentration of H⁺?and CH₃COO^-?is, therefore, the same

The equilibrium expression can be simplified to:

• Step 4:?Substitute the values into the expression to find?K_a

=?1.74 x 10^-5

• Step 5:?Determine the units of?K_a

= mol dm^-3

The value of?K_a?is therefore 1.74 x 10^-5?mol dm^-3

• Step 6:?Find pK_a

pK_a?= - log_10?K_a

= - log₁₀?(1.74 x 10^-5)

= 4.76

K_w

• The?K_w?is the?ionic product of water
  • It is the?equilibrium constant?for the dissociation of?water?at 298 K
  • Its value is 1.00 x 10^-14?mol²?dm^-6
• For the?ionisation?of water the equilibrium expression to find?K_w?is as follows:

H₂O (l) ? H⁺?(aq) + OH^-?(aq)

• As the?extent of ionisation?is very low, only small amounts of H⁺?and? OH^-?ions are formed
• The concentration of H₂O can therefore be regarded as constant and removed from the?K_w?expression
• The equilibrium expression therefore becomes:

K_w?= [H⁺] [OH^-]

• As the [H⁺] = [OH⁺] in pure water, the equilibrium expression can be further simplified to:

K_w?= [H⁺]²

Worked Example: Calculating the concentration of H⁺?of pure water

Answer

• Step 1:?Write down the equation for the partial dissociation of water

?In pure water, the following equilibrium exists:

H₂O (l) ? H⁺?(aq) + OH^-?(aq)

• Step 2:?Write down the equilibrium expression to find?K_w

• Step 3:?Simplify the expression

Since the concentration of H₂O is constant, this expression can be simplified to:

K_w?= [H⁺] [OH^-]

• Step 4:?Further simplify the expression

The ratio of H⁺?to OH^-?is 1:1

The concentration of H⁺?and OH^-?is, therefore, the same and the equilibrium expression can be further simplified to:

K_w?= [H⁺]²

• Step 5:?Rearrange the equation to find [H⁺]

K_w?= [H⁺]²

• Step 6:?Substitute the values into the expression to find?K_w

= 1.00 x 10^-7?mol dm^-3