What is the pH of the buffer formed by mixing 50.00 mL of 0.182 mol/L NaH2PO4 (Ka for H2PO4- = 6.32x10-8) with 50.00 mL of 0.112 mol/L NaOH.

Answer 1

NaH₂PO₄ ==> Na⁺ + H₂PO₄^- and H₂PO₄^-is a weak acid

H₂PO₄^- + NaOH ==> NaHPO₄^- + H₂O and NaHPO₄^- is the salt of that weak acid

THIS FORMS THE BUFFER

Initial moles H₂PO₄^- = 50.00 ml x 1 L/1000 ml x 0.183 mol/L = 0.00916 moles

Initial moles NaOH = 50.00 ml x 1 L/1000 ml x 0.112 mol/L = 0.0056

H₂PO₄^- + NaOH ==> NaHPO₄^- + H₂O

0.00916....0.0056............0...................Initial

-0.0056.....-0.0056.......+0.0056...........Change

0.00356......0.................0.0056............Equilibrium

We can use the Henderson Hasselbalch equation to find the pH

pH = pKa + log [salt]/[acid] but first we need to find the pKa from the given Ka

pK = -log Ka = -log 6.32x10^-8 = 7.2

pH = 7.2 + log (0.0056/0.00356) = 7.199 + 0.197

pH = 7.396 = 7.40