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AQA A Level Chemistry復習筆記5.6.1 Acid Dissociation Constant

Category: A-level課程, 教材筆記, 福利干貨 Date: 2022年6月27日 下午3:25

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

 

轉載自savemyexams

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