Question

Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.

Answer 1

Step-by-step explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that
`M_(1)V_(1)` =
`M_(2)V_(2)`

Where,
`V_(1)` = total volume

`V_(2)` = initial volume

Therefore, new concentration of
`CH_(3)COOH` =
`(M_(2)V_(2))/(V_(1))`

=
`(60 * 0.5)/(70)`

= 0.43 M

New concentration of NaOH =
`(M_(2)V_(2))/(V_(1))`

=
`(10 * 1.0)/(70)`

= 0.14 M

So, the given reaction will be as follows.

`CH_(3)COOH + OH^(-) \rightarrow CH_(3)COO^(-) + H_(2)O`

Initial: 0.43 0.14 0

Change: -0.14 -0.14 0.14

Equilibrium: 0.29 0 0.14

As it is known that value of
`pK_(a)` = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

pH =
`pK_(a) + log ([CH_(3)COO^(-)])/([CH_(3)COOH])`

=
`4.74 + log (0.14)/(0.29)`

= 4.74 + (-0.316)

= 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.