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50.50 mL of 0.116 M HF is titrated with 0.1200 M NaOH. What is the pH...

50.50 mL of 0.116 M HF is titrated with 0.1200 M NaOH. What is the pH when 25.00 mL of the base have been added. (Ka = 6.8 x 10-4)

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50.50 mL of 0.116 M HF is titrated with 0.1200 M NaOH. What is the pH when 25.00 mL of the base have been added. (Ka = 6.8 x 10-4)

    HF +       NaOH ----------------------> NaF            +    H2O

50.5 x 0.116             0.12 x 25                     0                     0

5.86                          3                                 0                         0      --------------initial millimoles

2.86                           0                                 3                       3        ----------------at equilibrium

the mixture contain NaF salt + HF acid so it is buffer

For acidic buffer

pKa = -log Ka = -log(6.8 x 10-4) = 3.167

pH = pKa + log[salt/acid]

pH = 3.167 + log[ 3 / 2.86]

pH= 3.19

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