Question

Prepare 100mL of a 0.1M potassium phosphate buffer of pH 6.5. You have K2HPO4 and KH2PO4 in solid/salt form available to mix with water to create this buffer. Final answer must include how many grams of each salt to mix with how many mL of water to prepare the buffer.

I started this problem and got as far as using the Henderson-Hasselbalch equation and that the ratio of [HPO42-] to [H2PO4-] is 0.48 moles to 1 mole.

Answer 1

Phosphoric acid pKa's are 2.3, 7.21 and 12.35.
Our required pH is 6.5, so 7.21 will be used.
Hence, monopotassium dihydrogen phosphate (KH2PO4) and dipotassium mono-hydrogen phosphate (K2HPO4) will be used.

Handerson Hasselbach equation:

pH = pKa + log ([A-]/[HA])

Therefore, A- is K2HPO4 concentration, and for HA is KH2PO4 concentration:

6.5 = 7.21 + log(A/HA)
log(A/HA)=-0.71
A/HA = 0.49

We know that total salt concentration = 0.1 M

HA+A=0.1

A/HA=0.49

HA+0.49 HA=0.1
[HA] = 0.1/1.49 = 0.067M
[A] = 0.1-0.067M = 0.033M

Therefore, the concentration of K2HPO4 = 0.033M
and the concetration of KH2PO4 = 0.067M

Since, we are preparing 100 mL or 0.1 L of the solution,
the wieght of K2HPO4 = (0.033M x mol.weight)x 0.1L
                   = (0.033M x 174.2 g/mol)x 0.1L
                   = 0.575 g of K2HPO4 required for 0.1L buffer solution.

                  
Similarly, we can calculate for KH2PO4:
                  
wieght of KH2PO4 = (0.067M x mol.weight)x 0.1L
               = (0.067M x 136.086 g/mol)x 0.1L
               = 0.912 g of KH2PO4 required for 0.1L buffer solution.
              
Therefore, we need to take 0.575 g of K₂HPO₄ + 0.912 g of KH₂PO₄ and dissolve them in 100 mL water to
prepare 100mL of a 0.1M potassium phosphate buffer of pH 6.5.