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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?

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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:

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

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}} &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90} = \pu{311.7 g}\\ m_{\mathrm{LA}} &= \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.

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You should be aware, that in concentrated solutions lactic acid polymerizes into mainly lactyllactic acid Watson (1940), which reduces the amount of “free” acid by around 20%. The polymers are sufficiently stable at room temperatures to create inconsistencies between what is actually necessary for neutralisation (titration) and the theoretical value. Therefore, analyses for titration of concentrated lactic acid prescribe boiling with excess of Na/K-hydroxide for about 20 minutes and a back-titration of the remaining OH.

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  • $\begingroup$ This is fascinating. Thank-you. $\endgroup$ – Dave Feb 26 at 13:36
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90 g pure lactic acid reacts with 56 g pure potassium hydroxyde. As your substances are not pure, you must use 102 g of your impure lactic acid (88%), and make it react with 62.2 g of your impure (90%) potassium hydroxyde.

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    $\begingroup$ It is unnecessary to post a new answer. Please edit an older one; if it is deleted and you cannot undelete it, please flag it for moderator assistance. $\endgroup$ – Martin - マーチン Feb 25 at 17:51
  • $\begingroup$ Thank you for your help! $\endgroup$ – Dave Feb 26 at 13:24

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