Table of Ka values
pKa = -logKa
Table of pKa values
Finding Ka and pKaAt 298 K, a solution of 0.100 mol dm-3 ethanoic acid has a hydrogen ion concentration of 1.32 x 10-3 mol dm-3.Calculate the Ka & pKa of the acid.
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 Ka
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 Ka
= 1.74 x 10-5
Step 5: Determine the units of Ka
= mol dm-3
The value of Ka is therefore 1.74 x 10-5 mol dm-3
Step 6: Find pKa
pKa = - log10Ka
= - log10 (1.74 x 10-5)
= 4.76
HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)
HA (aq) + OH- (aq) → A- (aq) + H2O (l)
Finding the Ka of a weak acid from a pH titration curve
You can regard the symbol p as meaning -log10 of a value. You don't need to include the 10 as 'log' means log base 10. If a natural logarithm (base e) is required it is given the symbol ln.Other uses of p include pOH and pKw . The latter gives a useful shortcut in problem solving:
Kw = [H+][OH-] = 1.00 x 10-14 mol2 dm-6 at 298 K-logKw = -log[H+] + (-log[OH-]) = -log(1.00 x 10-14)
pKw = pH + pOH = 14.00
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