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A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL of water...

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Chemistry

A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL ofwater and titrated with 0.41 M NaOH. After 33 mL of the NaOHsolution 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’sleft 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???

Answer & Explanation Solved by verified expert
4.4 Ratings (929 Votes)

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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