Question

Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at pH ~pKa D. 0.2 M carbonic/bicarbonate (H2CO3/HCO3-) at pH ~pKa

Answer 1

Explanation:

(A)   As we know that carbonic acid ($H_{2}CO_{3}$) and Sodium bicarbonate ($NaHCO_{3}$) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = $pK_{a} + log(\frac{[Salt]}{[Acid]})$ ........... (1)

So mathematically,  if [Salt] = [Acid]  then $\frac{[Salt]}{[Acid]}$ = 1 .

And,  $log (\frac{[Salt]}{[Acid]})$ = 0

Therefore, equation (1) gives us the following.

         pH = $pK_{a}$ (when acid and salt are equal in concentration)

Hence, $pK_{a}$ of $H_{2}CO_{3}$ (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = $pK_{a}$ = 6.35.

(B)    $[NaHCO_{3}]$ = 0.045 M and,  $[H_{2}CO_{3}]$ = 0.45 M

Hence,   pH = $6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])$

                     = $6.35 + log(\frac{0.045}{0.45})$

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    $[NaHCO_{3}]$ = 0.45 M $[H_{2}CO_{3}]$

                          = 0.045 M

So,       pH = $6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})$

                  = $6.35 + log(\frac{0.45}{0.045})$

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.