Question

Consider a buffer solution that contains 0.50 M NaH2PO4 and 0.20 M Na2HPO4. pKa(H2PO4-)=7.21. a) Calculate the pH. b)Calculate the change in pH if 0.120 g of solid NaOH is added to 150 mL of this solution. c) If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H3O+ can be neutralized by 250 mL of the initial buffer.

Answer 1

a) Calculate the pH.

pH = pKa + log(HPO4-2 / H2PO4-)

pH = 7.21 + log(0.2/0.5)

pH = 6.81

b)Calculate the change in pH if 0.120 g of solid NaOH is added to 150 mL of this solution.

mol of H2PO4- = MV = 0.5*150 = 75

mol of HPO4-2 = MV = 0.2*150 = 30

mmol of NaOH = mass/MW = (120/40) = 3

after reaction

mol of H2PO4- = 75 - 3 = 72

mol of HPO4-2 = 30 + 3 = 33

pH = pKa + log(HPO4-2 / H2PO4-)

pH = 7.21 + log(33/72)

pH= 6.87

c) If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H3O+ can be neutralized by 250 mL of the initial buffer.

pHmin = 6.81-0.10 = 6.71

mmol of HPO4-2 = MV = 0.2*250 = 50

mmol of H2PO4- = MV = 0.5*250 = 125

after "x" mol of H+ is added

mmol of HPO4-2 = 50 - x

mmol of H2PO4- = 125 + x

pH = pKa + log(HPO4-2 / H2PO4-)

6.71= 7.21 + log((50-x)/(125+x))

10^(6.71-7.21 ) = (50-x)/(125+x)

0.3162*125 + 0.3162x = 50-x

(1+0.3162)x = 50-0.3162*125

x = (50-0.3162*125 ) / ((1+0.3162)) =

x = 7.958 mmol

x = 7.958*10^-3 mol = 0.007958