Converting Lactic Acid to Potassium Lactate I am looking to try an experiment incorporating potassium lactate into soap making. This seems to be new territory for soap makers in that sodium lactate is usually used. I found a bottle of "Lactic Acid 88%" at a beer making supplier and already have 90% potassium hydroxide but do not know how much of one is needed to neutralize the other. Can anyone point me to the right tables or resources to do the calculations?
OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -&gt; CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$.  Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$: 

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid.  Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

\begin{align}
m_{\ce{KOH}} 
  &amp;= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}
 = \pu{311.7 g}\\
m_{\mathrm{LA}} 
  &amp;= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88}
 = \pu{511.8 g}
\end{align}

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.