# Adjusting the ph of salicylic acid using sodium hydroxide

Question
How many milliliters of $0.202\ \mathrm M$ NaOH should be added to $25.0\ \mathrm{mL}$ of $0.0233\ \mathrm M$ salicylic acid to adjust the $\mathrm{pH}$ to $3.50$?
The answer from the text book is $2.2\ \mathrm{mL}$ NaOH My attempt I believe the correct equation to use for this is the Henderson–Hasselbalch equation

$$\mathrm{pH}=\mathrm pK_\mathrm a + \log\left(\frac{[\ce B]}{[\ce{BH}]}\right)$$

$\mathrm pK_\mathrm a = 2.972$

So I set it up as:

$$3.50=2.972+\log\left(\frac{x}{0.0233}\right)$$

Solving for $x$ I get $0.0785$

I don't know where to go from here since $x$ is the concentration I figure that if I multiply by $0.025$ (the volume) and then divide by the molarity ($0.202$) I would get the answer but that's not right.

The Henderson-Hasselbalch equation is the correct equation, but ask yourself what you are putting in there: $$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$

Since $A^-$ and $HA$ occupy the same volume, you can rewrite: $$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$

When adding Sodium Hydroxide consider the following equation $$\ce{AH + NaOH <=> A- + H2O + Na+}$$

Given that you will find $$n(A^-) = n(\ce{NaOH})\\ n(HA) = n_0(HA)-n(\ce{NaOH})$$

Then you can substitute and solve for $n(\ce{NaOH})$: $$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$

You then divide by the concentration, yielding your volume.

Intermediate results:

Initial amount of $HA$:

$n_0(HA) = 0.5825\ \text{mmol}$

Amount of $\ce{NaOH}$:

$n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$

Final Volume:

$V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$