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I am not sure how to understand when pH will affect a given problem regarding extraction. For example, in a problem given that K$_D$ of the reaction is $1.86$ I am asked to calculate:

(a) How much propionic acid is left in octanol if you shake $100$ ml of octanol as contains $1.0$ mM propionic acid with a $50$ ml aqueous solution buffered to pH $2$?

(b) How much propionic acid is left in octanol if you shake $100$ ml of octanol as contains $1.0$ mM propionic acid with a $50$ ml aqueous solution buffered to pH $7$?

For these problems it is stated that "At pH $2$, only AH will be formed in water and we can ignore the proton equilibrium, so that n$_0$ = [AH]$_{aq}$ V$_{aq}$ + [AH]$_{org}$ V$_{org}$" but " At pH $7$, A- is no longer negligible and n$_0$ = [AH]$_{aq}$ V$_{aq}$ + [AH]$_{org}$ V$_{org}$ +[A-]$_{aq}$ V$_{aq}$"

How does one know this?

The same is goes for the problem:

Propionic anhydride can be extracted from cyclohexane to water. The distribution equilibrium can be written as:

PRANH (org) <-> PRANH (H2O) K = $0.82$

In aqueous solution, propionic anhydride can be hydrolyzed to propionic acid (PRIC ACID) according to the following equilibrium:

PRANH + H2O <-> 2 PRIC ACID K = $0.0023$ M (at pH <$3.0$)

A cyclohexane solution contains 1$0$ mM propionic anhydride. At what volume ratio of cyclohexane to water will $80$% of the propionic anhydride be extracted over to the aqueous phase (pH $2$) with a shake?

Here I assume that I should either get if the pH doesn't matter:

$1-p = \frac{1}{1+ K_D V_{aq}/V_{org}}$

Or if pH matters then n$_0$ = [PRANH]$_{aq}$ V$_{aq}$ + [PRANH]$_{org}$ V$_{org}$ +[PRIC ACID]$_{aq}$ V$_{aq}$ and I would get that:

$1-p = \frac{1}{1+ K_D* V_{aq}/V_{org}+ [PRIC ACID]_{aq}/[PRANH]_{org}*V_{aq}/V_{org}}$

where $1$-p=$1$-$0.8$=$0.2$

I would assume that the second is correct (if either is correct) since it is given that PRANH hydrolyzes at pH<$3$ so if we're at pH=$2$ then that should occur. But am I correct in assuming that? But then what happens to the equilibrium constant given for the hydrolyzation, do I use that for something or should I only take into account the K$_D$ = $0.82$ for the extraction process?

Thankful for all help!

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  • $\begingroup$ $\ce{HA(Oct)<=>HA(aq)<=>H+(aq) + A-(aq)}$. The former eq. is determined by KD, the latter by the acidity constant. What is unclear about that? Similarly for PRANH hydrolysis. $\endgroup$
    – Poutnik
    Commented Feb 19, 2022 at 11:44
  • $\begingroup$ @Poutnik How do I from the way that the question is stated know when to take into account the ions and when not to as in the first question? For the PRANH hydrolysis I assume that I should take into account the hydrolysis but is the way that I am applying this correct? I have just "copied" what I understood from the first question to the hydrolysis question.. $\endgroup$
    – katara
    Commented Feb 19, 2022 at 11:52
  • $\begingroup$ You use MathJax formatting in a weird way. // Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic $\endgroup$
    – Poutnik
    Commented Feb 19, 2022 at 11:57
  • $\begingroup$ Generally, components being extracted can undergo in one phase a reaction to a form inactive in extraction. Typically salts do not like to be ectracted to nonpolar solvents. The reactions often follow an equilibrium equation. From give equilibrium constants and eventually pH, there is calculated nonactive/total form ratio. If the ratio << 1, this equilibrium reaction can be neglected for the extraction evaluation. Otherwise it must be taken into consideration. $\endgroup$
    – Poutnik
    Commented Feb 19, 2022 at 12:09
  • $\begingroup$ Or, you can define conditional distribution constant [A_org]/[A_water, total], involving the mentioned ratio. $\endgroup$
    – Poutnik
    Commented Feb 19, 2022 at 13:15

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