# How to calculate the pH of a propanoate buffer after addition of sodium hydroxide?

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'.

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.