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
轉載自savemyexams
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