3
$\begingroup$

My book says that-

In acidic water, the negative fatty acid ions become fatty acids.

However, they did not include a chemical equation to demonstrate this, so I made my own:

$$\ce{C17H35COO- (aq) + HCl (aq) -> C17H35COOH (aq) + Cl- (aq)}$$

But according to this equation, the fatty acid ion is a base, since it gets a proton. So this must be wrong. Can anyone point out my mistakes (if any), and explain what happens when a negative fatty acid ion comes in contact with acidic water, and becomes a fatty acid?

$\endgroup$
1
  • $\begingroup$ Here's a thought: Try adding vinegar to a solution of soap and water. Will this be enough to break down all soap, and cancel out the effect of the soap molecules totally? Why / Why not? $\endgroup$
    – Lars Tuff
    Feb 24, 2016 at 2:57

2 Answers 2

4
$\begingroup$

Your equation is correct. Hydrochloric acid ($\ce{HCl}$) is a stronger acid than stearic acid ($\ce{C17H35COOH}$),i.e. it is dissociated into $\ce{H+}$ and $\ce{X-}$, the anionic rest, to a greater degree. Therefore, it protonates the stearate ion. One can also say that the conjugated base of stearic acid (the stearate ion) is a stronger base than chloride, the conjugated base of hydrochloric acid.

$\endgroup$
2
$\begingroup$

Actually, soap in acidic water just means reactions with the hydrogen ions in acid and whatever other element in the acids is not a matter to be concerned about. For example, sodium stearate $\ce{C17H35COONa}$ decomposes in in acidic water soap:

\begin{align} \ce{C17H35COONa &-> \underset{\text{(stearate ion)}}{C17H35COO-} + \underset{\text{(sodium ion)}}{Na+}} \\ \ce{C17H35COO- + \underset{\text{(from any acid)}}{H+} &-> \underset{\text{(stearic acid)}}{C17H35COOH}} \end{align}

Due to the formation of stearic acid, soap cannot act efficiently in acidic water.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.