Question

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buffers are 2.12, 7.21 and 12.67)?

Answer 1

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

               pH = $pK_{a} + log \frac{base}{acid}$

where,     pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So,      7.4 = 7.21 + $log \frac{base}{acid}$

             0.19 = $log \frac{base}{acid}$

            $\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

           $\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

  $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)        

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

             1.55$[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

                        2.55 $[H_{2}PO_{4}]$ = 1 mM

                             $[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

                     0.392 mM + $[HPO_{4}]$ = 1 mM

                          $[HPO_{4}]$ = (1 - 0.392) mM

                              $[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.