AQA A Level Chemistry复习笔记5.6.1 Acid Dissociation Constant

Acid Dissociation Constant

 

Weak acids

  • A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions
    • Eg. most organic acids (ethanoic acid), HCN (hydrocyanic acid), H2S (hydrogen sulfide) and H2CO3 (carbonic acid)

     

  • The position of the equilibrium is more over to the left and an equilibrium is established

1.7-Equilibria-Dissociation-of-a-Weak-Acid

The diagram shows the partial dissociation of a weak acid in aqueous solution

 

  • As this is an equilibrium we can write an equilibrium constant expression for the reaction

5.6.1-The-acid-dissociation-constant

  • This constant is called the acid dissociation constant, Ka, and has the units mol dm-3
  • Values of Ka are very small, for example for ethanoic acid Ka = 1.74 x 10-5 mol dm-3
  • 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 value of Ka indicates the extent of dissociation
    • The higher the value of Ka the more dissociated the acid and the stronger it is
    • The lower the value of Ka the weaker the acid

     

Worked Example

Writing Ka expressionsWrite the expression for the following acids:

  1. Benzoic acid, C6H5COOH
  2. Carbonic acid, H2CO3

Answer

5.6.1-WE1-Writing-Ka-Expressions

 

pH of Weak Acids

Weak acids

  • The pH of weak acids can be calculated when the following is known:
    • The concentration of the acid
    • The Ka value of the acid

     

  • From the Ka expression we can see that there are three variables:

5.6.1-The-acid-dissociation-constant

  • However, the equilibrium concentration of [H+] and [A-] will be the same since one molecule of HA dissociates into one of each ion
  • This means you can simplify and re-arrange the expression to

Ka x [HA] = [H+]2

[H+]2 = Ka x [HA]

  • Taking the square roots of each side

[H+] = √(Ka x [HA])

  • Then take the negative logs

pH = -log[H+] = -log√(Ka x [HA])

Worked Example

pH calculations of weak acidsCalculate the pH of 0.100 mol dm-3 ethanoic acid at 298 k with a Ka value of 1.74 × 10-5 mol dm-3

Answer

Ethanoic acid is a weak acid which ionises as follows:

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 1: Write down the equilibrium expression to find Ka

5.6.1-WE2-answer-1

Step 2: Simplify the expression

The ratio of H+ to CH3COO- ions is 1:1

The concentration of H+ and CH3COO- ions are therefore the same

The expression can be simplified to:

5.6.1-WE2-answer-2

Step 3: Rearrange the expression to find [H+]

5.6.1-WE2-answer-3-1

Step 4: Substitute the values into the expression to find [H+]

5.6.1-WE2-answer-4-1

= 1.32 x 10-3 mol dm-3

Step 5: Find the pH

pH = -log[H+]

= -log(1.32 x 10-3)

= 2.88

 

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