Question

You have 0.500 L of an 0.250 M acetate buffer solution (i.E. [HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M) at pH 3.50. How many mL of 1.000 M NaOH must you add in order to change the pH to 5.25? Acetic acid has a pKa of 4.74.

Answer 1

Explanation:

Relation between pH and $pK_{a}$ is as follows.

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

Hence, total moles of acetate is calculated as follows.

           $0.250 M \times 0.5 L$

           = 0.125 M

     $\frac{B}{A} = 10^{{pH} - pk_{a}}$

                 = $10^{3.50 - 4.74}$

                 = 0.0575

Let us assume that there are x moles of acid. Hence, moles of B are as follows.

                 B = (0.125 - x) moles

      $\frac{0.125 - x}{x}$ = 0.0575

                  x = 0.118 moles

Now, we will calculate the concentration as follows.

           5.25 = 4.74 + $log \frac{[B]}{[A]}$

           $log \frac{[B]}{[A]} = 10^{5.25 - 4.74}$

                        = 3.236

                     x = $\frac{0.125}{4.236}$

                         = 0.0295 moles

Therefore, moles of NaOH needed to add are as follows.

        $M_{NaOH} \times V_{NaOH}$ = (0.118 - 0.0295) moles

          $1.00 M \times V_{NaOH}$ = 0.0885 moles

                            = $0.0855 L \times \frac{10^{3}ml}{L}$

                            = 88.5 ml

Thus, we can conclude that there is 88.5 ml of 1.000 M NaOH must you add in order to change the pH to 5.25.