The diagram shows the partial dissociation of a weak acid in aqueous solution
Writing Ka expressionsWrite the expression for the following acids:
Answer
Ka x [HA] = [H+]2
[H+]2 = Ka x [HA]
[H+] = √(Ka x [HA])
pH = -log[H+] = -log√(Ka x [HA])
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
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:
Step 3: Rearrange the expression to find [H+]
Step 4: Substitute the values into the expression to find [H+]
= 1.32 x 10-3 mol dm-3
Step 5: Find the pH
pH = -log[H+]
= -log(1.32 x 10-3)
= 2.88
转载自savemyexams
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