# Finding pH of a buffer solution after addition of NaOH or HCl

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.

• 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. – M. Farooq Feb 11 at 20:16