139k views
5 votes
Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.

User Alfia
by
8.8k points

1 Answer

2 votes

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.

User Tsturzl
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.