Question

# A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH...

A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. Calculate the concentration of acetic acid in the 25.0 mL sample. Calculate the pH at the equivalence point (Ka acetic acid = 1.75x10^-5)

It is a weak acid- strong base titration

A ) At equivalence point : Moles of acid in sample = Moles of base added to sample

Moles of base added to sample = molarity * volume ( in Litre)

= 0.09984mol/litre * 0.0375 Litre = 3.744*10^-3 moles

Moles of acid in sample = 3.774*10^-3 moles

Molarity of sample = moles / volume (litre) = 3.774*10^-3/0.025 = 0.15 M

B) pH at equivalence point :

CH3COOH +    NaOH --> NaCH3COO +     H2O

Initial(moles)             3.774*10^-3   3.774*10^-3             0

Change                   -3.774*10^-3   -3.774*10^-3      +3.774*10^-3

Final                              0                            0             3.774*10^-3

Total volume will be sum of volume of CH3COOH and Volume of NaOH = 0.025 + 0.0375 = 0.0625L

Molarity of NaCH3COO = total moles/ total volume ( in litre) = 3.774*10^-3/0.0625 = 0.060M

Now This base NaCH3COO will be in equilibrium in its conjugate acid

NaCH3COO +     H2O -->    NaCH3COOH + OH-

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