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.

  • $\begingroup$ 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

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