# Equivalence point of titration

Why is $\ce{[HIn] = [In-]}$ at equivalence point of a titration? I know that at equivalence point moles of known solution is equal to the moles of unknown solution, but I'm not able to relate that fact with the above conclusion. Perhaps I'm don't know something.

• Please no mathjax in titles. meta.chemistry.stackexchange.com/questions/261/… – orthocresol Aug 27 '16 at 9:34
• @Shuvam at equivalence point the reductant and oxidant are equal in their mole stoichiometric. – JM97 Aug 27 '16 at 9:37
• @Shuvam hope this helps chemguide.co.uk/physical/acidbaseeqia/indicators.html – JM97 Aug 27 '16 at 9:56
• What is $\mathrm{HIn}$? The indicator? If so, this statement is false. The equivalence point is by definition when you have added a stoichiometric amount of the titrant to the analyte. – Zhe Oct 27 '16 at 17:35

The important choice in a titration is choosing the correct indicator. If you are titrating a strong acid with a strong base (so the titration’s equivalence point is approximately $7$), you want an indicator whose equivalence point is just behind that (at ca. $8$), so that the tiniest of drops added additionally will cause the colour change.
As you might already know, $\ce{[In^-]}$ is the concentration of the indicator that was added to the solution, which basically shows whether the solution is basic or acidic or neutral. The equivalence point of a titration is when the titrant has reacted with the unknown chemical solution so that it reaches an endpoint of the indicator. The equalization of the concentration $\ce{[HIn] = [In^-]}$ is when the color is at its neutral state: neither colors of the indicator is dominant over the other.