BUFFER SOLUTION P

Question

What is the pH of the buffer formed by mixing 50.00 mL of 0.205 mol/L NaH2PO4 (Ka for H2PO4- = 6.32x10-8) with 50.00 mL of 0.111 mol/L NaOH.

(Enter your answer to three significant figures. Ignore the dissociation of water.)

Answer 1

0.05000 L x 0.205 mol/L = 0.01025 moles NaH₂PO₄

0.05000 L x 0.111 mol/L = 0.00555 moles NaOH

NaH₂PO₄ + NaOH ===> NaHPO₄- + H2O

0.01025.......0.00555............0........................Initial

-0.00555.....-0.00555........+0.00555.............Change

0.0047..............0................0.00555..............Equilibrium

Using the Henderson Hasselbalch equation we have...

pH = pKa + log [NaHPO₄-]/[NaH₂PO₄]

Final volume = 50.00 + 50.00 ml = 100.00 ml = 0.10000 L

[NaHPO₄-] = 0.00555 mol/0.1 L = 0.0555 M

[NaH₂PO₄] = 0.00470 mol/0.1 L = 0.0470 M

pKa = -log Ka = -log 6.32x10^-8 = 7.20

pH = 7.20 + log (0.0555/0.0470)

pH = 7.20 + 0.072

pH = 7.27