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I've calculated the pH of the buffer solution, as seen. The question I'm referring to is part d), which asks: 'Calculate the pH of the solution after 0.01 mols of NaOH are added to 500$cm^3$ of the solution'.

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The addition of the NaOH will "use up" 0.01 mol of propanoic acid and "create" 0.01 mol of propionate. To make things easier, just work in molarity. Thus 0.02 M NaOH.

The pH of the final solution can be found by adjusting the equation you used in part b,

$pH = -log \left( 1.26 \times 10^{-5}\frac{0.1-0.02}{0.05+0.02} \right)=4.84$

The result agrees with our expectations. Since the solution is a buffer, the pH is not expected to drastically change with the addition of a strong base. It is buffered.

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