A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL of water and titra
ID: 882230 • Letter: A
Question
A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL of water and titrated with 0.41 M NaOH. After 33 mL of the NaOH solution has been added,
the overall pH = 5.281. Calculate the Ka value for the HA.
Let’s write out the acid dissociation:
HA A– + H+ Ka=???
Now, if we add NaOH (strong base), it will react with the HA (acid). So let’s construct an ICE table in moles to find out what’s left after reaction:
HA + OH– H2O + A–
I
0.035
0.0135
—
0
C
- 0.0135
- 0.0135
—
+0.0135
E
0.0215
0
—
0.0135
Can you pleae explain how they got the Change value== .0135???
HA + OH– H2O + A–
I
0.035
0.0135
—
0
C
- 0.0135
- 0.0135
—
+0.0135
E
0.0215
0
—
0.0135
Explanation / Answer
Let’s write out the acid dissociation:
HA A– + H+
Ka= [A-] [H+] / [HA]
The inital moles of acid = 0.035 moles
Moles of base added = moalrity X volume = 0.41 X 0.033 = 0.0135 moles
this much base will neutralize 0.0135 moles of acid
The equation of acid and base will be
HA + OH- --> H2O + A-
Initial mole 0.035 0.0135 0
Change -0.0135 -0.0135 0.0135
Equilibrium 0.0215 0 0.0135
pH = pKa + log [salt /acid]
5.281 = pKa + log [0.0135 / 0.0215]
Pka = 5.281 - log(0.627) = 5.281 - (-0.202) = 5.483
pKa = -logKa
taking antilog
Ka = 3.28 X 10^-6
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