A 1.30 L buffer solution consists of 0.107 M butanoic acid and 0.345 M sodium butanoate. Calculate the pH of the solution following the addition of 0.075 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

The pH of the solution following the addition of 0.075 moles of NaOH is 5.7.

Explanation:

The equation of the buffer solution is the following:

C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq)     (1)    

The pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

  (2)

The NaOH added will react with butanoic acid tot produce sodium butanoate:

C₄H₇O₂H(aq) + OH⁻(aq) ⇄ C₄H₇O₂⁻(aq) + H₂O(l)  (3)          

Now, from equation (1) we have:

C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq)

0.049 - x                           0.403 + x            x  

By solving the above equation for x we have:            

x = 1.85x10⁻⁶ = [H₃O⁺]

So, the concentration of butanoic acid and sodium butanoate is:

Finally, from equation (2) we have:

Therefore, the pH of the solution following the addition of 0.075 moles of NaOH is 5.7.

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