-1
$\begingroup$

I have a question asking me to find the pH of a phosphate buffer solution that has $\pu{80.0mM}$ of a buffer at $\pu{pH 6.8}$, after the addition of varying moles of $\ce{NaOH}$ or $\ce{HCl}$. The $\ce{pK_A}$ is $\pu{6.8}$.

One question, for $\ce{NaOH}$, was the addition of $\pu{0.016 moles}$ of $\ce{NaOH}$.

What I did for this was first find the concentrations of both HA and A, which was $\pu{0.04M}$, based off of the initial information.

Next, I proceeded to calculate the changes, with

$$HA = \pu{0.04M x 0.4L - 0.016 moles = 0}$$

$$A = \pu{0.04M x 0.4L + 0.016 moles = 0.032}$$

However, the existence of that 0 throws me off, as I can't throw that into the Henderson-Hasselbalch equation, for $log(0)$...

For the $\ce{HCl}$ addition, it is adding $\pu{0.024 moles}$. I do the same, but one of the results is

$$\pu{0.04M x 0.4L - 0.024 = -0.008}$$ where I can't apply $log$ to a negative number.

I'm pretty sure I've gone off track somewhere in my line of thinking. Any help would be appreciated! Thanks.

$\endgroup$
  • $\begingroup$ please search online on calculating buffer pH of a phosphate system. The hint is that phosphate has three acidic protons. The addition of NaOH or HCl is not going to consume all. $\endgroup$ – M. Farooq Feb 11 at 20:16

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.