Question

How many milliliters of 1.92 M NaOH must be added to 1.00 × 102 mL of 0.16 M NaH2PO4 to make a buffer solution with a pH of 6.50?

_____________ml

Answer 1

Let x be the volume of NaOH

Moles of NaOH = x / 1000 x 1.92 = 0.00192 x mol

Initial moles of H2PO4- = 1.0 x 10^2 / 1000 x 0.16 = 0.016 mol

NaOH + H2PO4- => HPO42- + Na+ + H2O

Moles of H2PO4- left = 0.016 - 0.00192 x mol

Moles of HPO42- = moles of NaOH = 0.00192 x mol

pH = pKa + log([HPO42-]/[H2PO4-])

pH = pKa + log(moles of HPO42- / moles of H2PO4-)

H3PO4Ka2 = 6.2×10^(–8) pKa2 = -logKa2 =-log(Ka2 = 6.2×10^(–8)) = 7.21

required pH = 6.50

6.50= 7.21 + log(0.00192x / (0.016- 0.00192x))

log(0.00192x / (0.016 - 0.00192x)) = -0.71

0.00192x / (0.016 - 0.00192x) =10^(-0.71) = 0.194

x = 1.353

Volume of NaOH = a = 1.353 mL (approximately 1.35 mL)

Thank you