Assuming the pH of a buffer solution that contains 0.65 M NaH2PO4 and 0.11M Na2HPO4 is 6.44.
Calculate the change in pH if 0.070 g of solid NaOH is added to 100 mL of the solution in the problem above
Answer:
∆ pH = 0.08
Explanation:
Calculate concentration in mmoles of each species before adding the NaOH.
[NaH2PO4] = M NaH2PO4 x mL NaH2PO4 = (0.65)(100) = 65 mmoles NaH2PO4
[Na2HPO4] = M Na2HPO4 x mL Na2HPO4 = (0.11)(100) = 11 mmoles Na2HPO4
When you add the strong base NaOH to the buffer, it will react with the acid part of the buffer (i.e. the species with the most Hs --- NaH2PO4).
0.070 g NaOH x (1000 mg / g) x (1 mmole NaOH / 40.00 mg NaOH) = 1.75 mmoles NaOH
[Concentration] . . . . . .NaH2PO4 + NaOH ➡ Na2HPO4 + H2O
initial . . . . . . . . . . 65 . . . . . .1.75 . . . . . . . . .11
change . . . . . . . .-1.75 . . . . .-1.75 . . . . . . .+1.25
final . . . . . . . . . . .63 . . . . . . . .0 . . . . . . . . . .13
pKa for H2PO4- = 6.2 x 10^-8; pKa = 7.21
pH = pKa + log ( [Na2HPO4] /[NaH2PO4]) = 7.21 + log (13/ 63) = 7.21 - 0.69 = 6.52
∆pH is minute (from 6.44 to 6.52). = 0.08
