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

Step-by-step explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and
`pK_(a)` is as follows.

pH =
`pK_(a) + log (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 (base)/(acid)`

0.19 =
`log (base)/(acid)`

`(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.

`(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.