# In this reaction: "HIn" (aq) rightleftharpoons "H"^(+)(aq) + "In"^(-)(aq) What happens when these stressors are added? Does the stress add or remove products or reactants? What's the shift? Which ion is involved in the reaction?

May 25, 2017

This is what will happen.

#### Explanation:

We usually use $\text{HIn}$ to represent a weak acid, which, as its name suggests, does not ionize completely in aqueous solution, that acts as a $\text{pH}$ indicator.

${\text{HIn"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "In}}_{\left(a q\right)}^{-}$

The unionized $\text{HIn}$ and the ionized ${\text{In}}^{-}$, the conjugate base of the weak acid, will have specific colors in aqueous solution.

The ionization equilibrium that is established in solution is governed by Le Chatelier's Principle, which states that the system will react to any stress placed on the current position of the equilibrium by shifting in such a way as to reduce that stress.

Now, hydrochloric acid is a strong acid, so it ionizes completely in aqueous solution

${\text{HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

When you're adding hydrochloric acid to your indicator, the excess hydrogen ions will cause the equilibrium to shift to the left.

${\text{HIn"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "In}}_{\left(a q\right)}^{-}$

color(white)(aaa)stackrel(color(blue)(larr))(color(white)(aaacolor(red)("shift to the left")aaaa)

This happens because the reverse reaction

${\text{H"_ ((aq))^(+) + "In"_ ((aq))^(-) -> "HIn}}_{\left(a q\right)}$

will consume some of the added hydrogen ions, i.e. it will reduce the stress placed on the position of the equilibrium by the increase in the concentration of hydrogen ions.

When sodium hydroxide, a strong base, is added to the solution, the hydroxide anions will neutralize some of the hydrogen ions present in the indicator solution

${\text{OH"_ ((aq))^(-) + "H"_ ((aq))^(+) -> "H"_ 2"O}}_{\left(l\right)}$

As a result, the equilibrium will shift to the right.

${\text{HIn"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "In}}_{\left(a q\right)}^{-}$

color(white)(aaa)stackrel(color(blue)(rarr))(color(white)(aaacolor(red)("shift to the right")aaaa)

This happens because the forward reaction

${\text{HIn"_ ((aq)) -> "H"_ ((aq))^(+) + "In}}_{\left(a q\right)}^{-}$

will produce more hydrogen ions, i.e. it will reduce the stress placed on the position of the equilibrium by the decrease in the concentration of hydrogen ions.