Calculate the pH of a solution made from 0.0473 mol NaOH added to 1.00L of pure water. Assume no volume changes.

Question

Calculate the pH of a solution made from 0.0473 mol NaOH added to 1.00L of a buffer containing 0.116 mol each of a weak acid (pka=4.66) and its conjugate base. Assume no volume changes.

Answer 1

NaOH ==> Na+ + OH-

0.0473 M NaOH = 0.0473 M OH-

pOH = -log [OH-] = -log 0.0473

pOH = 1.33

pH = 14 - pOH = 14 - 1.33

pH = 12.67

Let the weak acid be HA

HA + NaOH ==> NaA + H2O

0.116....0.0473.........0.116..................Initial

-0.0473..-0.0473....+0.0472...............Change

0.0687.......0............0.1633................Equilibrium

Henderson Hasselbalch equation:

pH = pKa + log [base] / [acid]

pH = 4.66 + log (0.1633 / 0.0687)

pH = 4.66 + 0.376

pH = 5.04