Question

17. A phosphate-based buffer was prepared by dissolving 28.4 g of Na2HPO4 and 12.0 g of NaH2PO4 in one liter of solution. 
a) Calculate buffer pH. 
b) Calculate the pH of the solution by adding 5.0 mmol of HCl to 100 mL of the buffer.

Answer 1

17)

a)

the pH of the buffer solution is calculated using the following equation

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

A- = Na2HPO4

Na2HPO4 = 28.4 g

moles of Na2HPO4 = given mass / molar mass = 28.4/141.96 =  0.2 mol

[Na2HPO4] = moles / volume in liter = 0.2/1 = 0.2 M

HA = NaH2PO4

NaH2PO4 = 12.0 g

moles of NaH2PO4 = 12 / 119.98 = 0.1 mol  

[NaH2PO4] = 0.1/1 = 0.1 M

pKa of H2PO4- = 7.2

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

pH = 7.2 + log ( 0.2 / 0.1)

pH = 7.2 + log (2)

pH = 7.2 + 0.3

pH = 7.5

pH of the buffer solution is = 7.5

b)

after addition of HCl to the solution, the following reaction takes place

H+ + HPO42- -----> H2PO4-

so the moles of H2PO4- increases and HPO42- decreases corresponding to the moles of HCl added

since 1 mole of HCl reacted with 1 mole of HPO42- produce 1 mole of H2PO4-

moles of HPO42- in 100 mL = molarity * volume in liter = 0.2 *100 = 20 mmol

moles of H2PO4- in 100 mL = 0.1*100 = 10 mmol

  

H+ + HPO42- -----> H2PO4-

initially 5 mmol 20 mmol 10 mmol

equilibrium 5 - 5 20 - 5 10+5

0 15 15

so

moles of HPO42- = A- = 15 mmol

moles of H2PO4- = HA = 15 mmol

pH = pKa + log (A-/HA)

pH = 7.2 + log ( 15/15)

pH = 7.2 + log 1

pH = 7.2 + 0

pH = 7.2

so the pH after addition of HCl = 7.2