Solution to PS 1 #10

To solve this problem, use the Henderson Hassalbalch equation with pH=7.0 and pKa=7.2

Let:

x= liters of K₂HPO₄

y= liters of KH₂PO₄

Then we know that x+y=1, therefore y=1-x.

Usng the definition of concentration we also know that:

[HPO₄^2-]=(0.01M)x/1Liter

[H₂PO₄^-]=(0.01M)y/1 Liter

So putting everything in The Henderson-Hassalbalch equation we get:

7.0=7.2+log{x/(1-x)}

Solving:

-0.2=log{x/(1-x)}

10^-0.2=x/(1-x)

0.631=x/(1-x)

x=0.631/1.631=0.387

y=1-x=1-0.387=0.613

Answer: Mix together 387 mls of 0.01M HPO₄² and 613 mls of 0.01M H₂PO₄^-.