# Finding new pH after NaOH added to buffer solution

I have a buffer containing 0.2 M of the acid $\ce{HA}$, and 0.15 M of its conjugate base $\ce{A-}$, with a pH of 3.35. I need to find the pH after 0.0015 mol of $\ce{NaOH}$ is added to 0.5 L of the solution.

I started by converting everything to moles, writing out an equation for $\ce{HA}$ and $\ce{OH-}$ reacting to make $\ce{H2O}$ and $\ce{A-}$, getting 0.0985 mol of $\ce{HA}$ left after all the $\ce{OH-}$ was used up, and 0.015 mol of $\ce{A-}$ produced.

At that point I got a bit stuck. I tried doing things a bunch of different ways after that, such as trying to calculate $\mathrm{p}K\mathrm{_a}$ by using the Henderson-Hasselbalch equation, but I'm not getting anywhere near the answer. I calculated a $\mathrm{p}K\mathrm{_a}$ of 3.47. I added the amount of $\ce{A-}$ made when $\ce{HA}$ reacts with $\ce{OH-}$ to the original moles of $\ce{A-}$ in the buffer (I calculated 0.18 M $\ce{A-}$), and tried putting that, the leftover moles of $\ce{HA}$, and my $\mathrm{p}K\mathrm{_a}$ value into the equation, but that wasn't right either.

What am I doing wrong?

• did you account for the change in volume? – bon Mar 15 '15 at 21:46
• Hmm no. Would I just assume a base of 1L originally for the plan buffer solution? I don't have a given initial amount. – Caesium-133 Mar 15 '15 at 21:49
• you said you have 0.5L of the buffer at the start – bon Mar 15 '15 at 21:53
• Ok, right, yea, sorry. So short of googling density of NaOH which would be odd since the book doens't usually give you questions where you'd have too look up something outside of the book, how would I figure out the change in volume? In the other questions it gave me a volume of NaOH for each thing. – Caesium-133 Mar 15 '15 at 21:55
• what is the final pH supposed to be? – bon Mar 17 '15 at 18:30

Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$

Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$

Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$

the Henderson Hasselbach equation is actually pH = pKa + log[A-]/[HA]

so using original conditions

3.35 = pKa + log[0.15/0.2]

Therefore pKa is 3.22

The rest of the equation is correct