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

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 $CH_{3}COOH$ = $\frac{M_{2}V_{2}}{V_{1}}$

                                        = $\frac{60 \times 0.5}{70}$

                                        = 0.43 M

New concentration of NaOH = $\frac{M_{2}V_{2}}{V_{1}}$

                                               = $\frac{10 \times 1.0}{70}$

                                               = 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 = $pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

                 = $4.74 + log \frac{0.14}{0.29}$

                 = 4.74 + (-0.316)

                 = 4.42

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