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A liter of a solution contains $\pu{0.120 mol}$ of $\ce{C6H5COOH}$ $(K_a = 6.5 ×10^{-5})$ and $\pu{0.105 mol}$ of $\ce{C6H5COONa}$. I'm asked to calculate the pH of the solution after adding $\pu{100 mL}$ of a $\pu{1.20 M}$ $\ce{NaOH}$ solution, the pH of the initial solution is $4.13$. Here's what I've done. Since $\ce{NaOH}$ is a strong base it will disassociate completely, so we will end up with $0.120$ moles of $\ce{OH^{-}}$. Calulating the $\ce{pOH}$ we get $0.92$ and with that we can calculate the $\ce{pH}$ which will be $13.07$. Is that the correct way of doing it ?

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Nope.

While $\ce{NaOH}$ is a strong base, the mixture of $\ce{C6H5COOH}$ and $\ce{C6H5COONa}$ is a buffer solution and tends to resist large changes in pH.

After adding the $\ce{NaOH}$, the extra $\ce{OH-}$ ions would react with the undissociated $\ce{C6H5COOH}$, effectively being removed from the solution. The solution would now be $1.1 \text{L}$ and contain a bunch of ions, like $\ce{Na+}$, $\ce{C6H5COO-}$, $\ce{H+}$, and $\ce{OH-}$. You can calculate the value of $\ce{[Na+]}$ now.

Use the Henderson's buffer equation:

$$\text{pH}=\text{p}K_a + \log{\left(\frac{\ce{[C6H5COO^-] }}{\ce{[C6H5COOH]}}\right)}$$

I think you can continue from here.

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  • $\begingroup$ I made a little big mistake earlier. @Buttonwood pointed out in chat that I wrote $\ce{CH3}...$ instead of $\ce{C6H5}...$. It's fixed now. $\endgroup$ May 28, 2017 at 15:01

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