0
$\begingroup$

The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of$ \ce{ KCl, CH_3 (OH) }$and $ \ce{CH3(CH2)_{11} O SO_3^{-} Na^+}$ at room temperature

enter image description here

What is the correct assignment(s) of sketches?

It is understandable that $ \ce{CH3(CH2)_{11} O SO_3^{-} Na^+}$ will have a sharp reduction surface tension with concentration (reference) and $\ce{KCl}$ increases surface tension with concentration since it is an inorganic salt. Through option elimination, we conclude that I must be methanol but would there be an independent way to conclude the same (without option elimination)?

I found this but the answers are not really authoritative.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buck Thorn
    Aug 23 '21 at 8:01
2
$\begingroup$

Curve I could be that of a detergent, but sodium dodecyl sufate decreases the tension faster than methanol and displays a very distinct change in behavior with onset of micellization at the critical micelle concentration. Above that concentration the amount of free solute - the concentration of which is accountable for the change in surface tension - remains approximately constant.

Methanol on the other hand is not a detergent and does not micellize. It does however reduce the surface tension of water, presumably because it disrupts the hydrogen bonding network in water. The boiling point of water is similarly decreased by addition of methanol.

Note that a similar decrease is observed for ethanol solutions, but that in some regards the ethanol-water system is more complicated (forming for instance an azeotrope).

Beyond this it is not possible to generalize, as hydrogen bonding renders water a very complicated liquid. There are something called structure forming (kosmotropic) and structure destroying (chaotropic) solutes. The ions dissociating when $\ce{KCl}$ dissolves jointly appear to have a net kosmotropic effect, but you'd have to distinguish this from the boiling point elevation expected on statistical grounds (Raoult's law). For any non-associating (eg associating by H-bonding) liquid adding a solute is expected to raise the boiling point. But in the case of water the effect of adding a solute might be that you enhance H-bonding. So it can be an effect beyond the statistical one.

$\endgroup$
5
  • $\begingroup$ If it disrupts and hence decreases, then can't we argue the same for KCL since the ions interrupt the hydrogen bonding? $\endgroup$
    – Buraian
    Aug 23 '21 at 14:09
  • $\begingroup$ Well, there is no simple answer. There are something called structure forming (kosmotropic) and structure destroying (chaotropic) solutes. The ions dissociating when KCl dissolves jointly have a net kosmotropic effect, and you'd have to distinguish this from the boiling point elevation expected on statistical grounds (Raoult's law). $\endgroup$
    – Buck Thorn
    Aug 23 '21 at 14:18
  • $\begingroup$ Where does raoult law come in here?) $\endgroup$
    – Buraian
    Aug 23 '21 at 14:19
  • $\begingroup$ Because for any non-associating (eg H-bonding) liquid adding a solute is expected to raise the boiling point. But in this case the effect might be that you enhance H-bonding. So it is an effect beyond the statistical one. $\endgroup$
    – Buck Thorn
    Aug 23 '21 at 14:20
  • 1
    $\begingroup$ Oh , so an negative deviating azeotrope. I see, this question is more complex than I had thought initially. $\endgroup$
    – Buraian
    Aug 23 '21 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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