Question

A chemist titrates 216.8 mL of a 0.3200 M butanoic acid (HC3H7CO2) solution with a 0.3200 M solution of NaOH. The titration setup shows that the butanoic acid solution is in the (Erlenmeyer) flask, and the NaOH solution is in the burette. The pKa of butanoic acid is 4.82. Calculate the pH of the solution in the flask after the chemist has added 231.4 mL of NaOH solution to it.

Answer 1

Answer: The pH of the solution in the flask after the chemist has added 231.4 mL of NaOH solution to it is 12.018.

Step-by-step explanation:

The given data is as follows.

Molarity of NaOH = 0.32 M

Molarity of
`HC_(3)H_(7)CO_(2)` = 0.32 M,

Volume of
`HC_(3)H_(7)CO_(2)` = 216.8 ml = 0.2168 L (as 1 L = 1000 mL)

Volume of NaOH = 231.4 ml = 0.2314 L

Therefore, number of moles of
`HC_(3)H_(7)CO_(2)` is as follows.

No. of moles = Molarity × Volume

=
`0.32 * 0.2168`

= 0.069 moles

Now, number of moles of NaOH will be calculated as follows.

No. of moles = Molarity × Volume

=
`0.32 * 0.2314`

= 0.074 moles

Chemical equation for this reaction is as follows.

`HC_(3)H_(7)CO_(2) + NaOH \rightarrow NaC_(3)H_(7)CO_(2) + H_(2)O`

Initial: 0.069 0.074

Change: -0.069 -0.069

Equilibm: 0 0.004672

Total volume will be as follows.

0.2168 + 0.2314 = 0.4482 L

Here, the concentration of NaOH is equal to the concentration of hydroxide ions.

So, [NaOH] =
`[OH^(-)] = (n)/(V)`

=
`(0.004672)/(0.4482)`

= 0.010424 M

We know that,

pOH =
`-log [OH^(-)]`

= -log (0.010424)

= 1.928

Hence, the pH will be calculated as follows.

pH = 14 - pOH

= 14 - 1.982

= 12.018

Thus, we can conclude that the pH of the solution in the flask after the chemist has added 231.4 mL of NaOH solution to it is 12.018.