In: Chemistry
A buffer solution is made by dissolving 0.230 moles of sodium hypochlorite in 1.00 L of 0.220 M hypochlorous acid (HClO). What is the pH of the solution after 0.0100 moles of sodium hydroxide is added? (The Ka of HClO is 2.9
Moles of HClO = M * V = 0.220 M * 1 L = 0.220 mol
Moles of NaClO = 0.230 mol
HClO + NaOH ------------> NaClO + H2O
0.220 0.01 0.230 ( Initial)
- 0.01 - 0.01 + 0.01 (change)
0.21 0 0.24 (Final)
Now,
pH before addition of NaOH is
pH = pKa + log( [ NaClO] / [HClO] )
pH = - log( 2.9 x 10^-8) + log(0.230 / 0.220)
pH = 7.537 + 0.02 = 7.56
pH after addition of NaOH is
pH = 7.56 + log[ ( NaClO)left / (HClO)left ]
pH = 7.56 + log( 0.24 / 0.21)
pH = 7.56 + 0.058 = 7.618 = 7.62