Question

In: Chemistry

A buffer solution is made by dissolving 0.230 moles of sodium hypochlorite in 1.00 L of...

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

Solutions

Expert Solution

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


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