-1
$\begingroup$

enter image description here

I am having trouble understanding the equations....how is concenteration of ester proportional to the volume of NaOH used and how is initial concentration proportional to difference in the Volume of NaOH used at infinity and time t=0....How did we find this relation

The same thing happens during the hydrolysis of sucrose or inversion of cane sugar where the reaction is monitored by polarimeter and the readings of the polarimeter are used .What is the common link in these types of reactions as both give equations in same type

Improve this question
$\endgroup$
1
  • $\begingroup$ Is the question alright now?? I didn't mean to ask this as a homework type format rather as one asking for the explanation for one of the steps here. $\endgroup$ – Aladdin Aug 1 '18 at 16:51
2
$\begingroup$

In a first order reaction it is only necessary to measure a quantity proportional to concentration. This could be a voltage or volume of reactant added etc. The reason this happens is because the ratio of quantities is all that is needed, thus if $X$ is some quantity at time $t$ and $X_0$ that initially there ($t=0$) then the equation for reaction is $\displaystyle X=X_0e^{-kt}$ where $k$ is the rate constant. Rearranging gives $\displaystyle \ln\left(\frac{X}{X_0}\right)=-kt$ and as you can see only the ratio is needed.

In your case, you can see from the equations you show, how the volume is used to be equivalent to something proportional to concentration $X$ and $X_0$ as used above. This is because of the way the titration works to neutralise acid with alkali (or vice versa) with equal amounts of each and for this an indicator is used to indicate the 'end point'.

Improve this answer
$\endgroup$
2
  • $\begingroup$ But I don't undrrstand one thing ......if difference between the volume at t=infinite and initial volume is directly proportional to a then why can't we say difference between volumes at t=t and volume at t=0 is proportional to a-x?.I am asking this silly doubt because I don't wanna cram the equation....instead learn how to use it $\endgroup$ – Aladdin Jul 31 '18 at 22:54
  • $\begingroup$ And what is the role of H+ ions here.Will they be come during rate lawv or are they ignored due to being in excess? $\endgroup$ – Aladdin Jul 31 '18 at 23:06

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