Answer to Question #312214 in Biochemistry for boo

Question

A 1 liter solution is made by adding 0.6209 moles NaH2PO4 and 0.8581 moles Na2HPO4. What will be the final pH of this solution after addition of 1. 9717 ml of 3. 8914 molar HCl? The pka of sodium dihydrogen phosphate is 6.86.

Answer 1

A 1 liter solution is made by adding 0.6209 moles NaH2PO4 and 0.8581 moles Na2HPO4.

[NaH2PO4] = 0.6209 mol/1 L = 0.6209 M

[Na2HPO4] = 0.1085 mol/1 L = 0.1085 M pKa of NaH2PO4 = 6.86 

After addition of 1.9717 mL of 3.8914 molar HCl. 

pOH= -log(1,

pH= 1.9$\times$10^-3

pH= 2.867