Question

Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium acetate and 1.00 M in acetic acid.

Answer 1

Explanation:

It is known that $pK_{a}$ value of acetic acid is 4.74. And, relation between pH and $pK_{a}$ is as follows.

                    pH = pK_{a} + log $\frac{[CH_{3}COOH]}{[CH_{3}COONa]}$

                          = 4.74 + log $\frac{1.00}{1.00}$

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  $CH_{3}COOH$ = moles of $CH_{3}COONa$

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            $CH_{3}COONa + NaOH \rightarrow CH_{3}COOH$

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log $\frac{[CH_{3}COONa]}{[CH_{3}COOH]}$

               = 4.74 +  log $\frac{0.94604}{1.05396}$

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.