Question

Hypothesize how your two buffers’ pHs will compare to each other. (Will one be higher than the other? If so, which one?) Explain.

 One solution will need 28-32 mL of NaH2PO4 and 18-22 mL Na2HPO4.

 The other will need 18-22 mL of NaH2PO4 and 28-32 mL Na2HPO4.

Assume you make a buffer with 20 mL KH2PO4 and 30 mL K2HPO4. Both solutions are provided at a concentration of 0.2000 M and are made to a final volume of 0.1000 L. Calculate the pH of this buffer. Show your work.

Answer 1

(a)

The acidic ion H2PO4- has a pKa value of 7.2

It dissociates to produce a proton and the conjugate base HPO4^2-.

Buffer pH is calculated using the Henderson Hasselbach equation, as shown below:

pH = pKa + log([conj. base]/[acid])

In this case, NaH2PO4 is the acid and Na2HPO4 is the conjugate base.

For the solution containing around 30 mL acid and around 20 ml conj. base, we have:

pH = 7.2 + log(20/30) = 7.2-0.176

For the solution containing around 20 mL acid and around 30 ml conj. base, we have:

pH = 7.2 + log(30/20) = 7.2+0.176

Thus, both these buffers have pH values within the buffering range, which is usually one unit on either side of the pKa value.

Also since both pH values lie symmetrically opposite to the pKa value, so both buffers are equally good.

(b)

In this case, we take 20 mL of the acid KH2PO4 and make the final volume upto 100 mL.

So, final conc of acid in solution = [acid] = Initial conc./Dilution factor = 0.2/(100/20) = 0.04 M

Also, we take 30 mL of the conj. base K2HPO4 and make the final volume upto 100 mL.

So, final conc of acid in solution = [acid] = Initial conc./Dilution factor = 0.2/(100/30) = 0.06 M

Putting the values in the Henderson Hasselbach equation:

pH = pKa + log([conj. base]/[acid]) = 7.2 + log(0.06/0.04) = 7.376