# Question 1f142

Mar 31, 2015

An indicator is a weak acid or a weak base that has different colors in undissociated and dissociated states.

In your case, methyl orange is a weak acid that's red in solutions that have a pH lower than 3.1 and yellow in solutions that have a pH larger than 4.4.

Because hydrochloric acid is a strong acid, it will dissociate completely to form

$H C {l}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

This means that the concentration of the ${H}^{+}$ ions will be equal to that of the $H C l$

$\left[{H}^{+}\right] = \text{0.05 M}$

Now, the generic equilibrium reaction for methyl orange (or any weak acid indicator, for that matter) is

$H I {n}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + I {n}_{\left(a q\right)}^{-}$

The equilibrium constant is defined as

${K}_{a} = \frac{\left[{H}^{+}\right] \cdot \left[I {n}^{-}\right]}{\left[H I n\right]}$

Use the pKa value given for methyl orange to determine the value of ${K}_{a}$

${K}_{a} = {10}^{- p {K}_{a}} = {10}^{- 3.7} = 1.99 \cdot {10}^{- 4} \cong 2.0 \cdot {10}^{- 4}$

Since you know $\left[{H}^{+}\right]$ and ${K}_{a}$, use them in the above equation

${K}_{a} = \frac{\left[{H}^{+}\right] \cdot \left[I {n}^{-}\right]}{\left[H I n\right]} \implies \frac{\left[I {n}^{-}\right]}{\left[H I n\right]} = {K}_{a} / \left(\left[{H}^{+}\right]\right)$

$\frac{\left[I {n}^{-}\right]}{\left[H I n\right]} = \frac{2.0 \cdot {10}^{- 4}}{0.05} = 4.0 \cdot {10}^{- 3}$

This is how much of the methyl orange will be left in basic form (yellow)

"%yellow" = ([In^(-)])/([HIn]) * 100 = 4.0 * 10^(-3) * 100 = color(green)("0.4%")

This means that the acid form (red) will be

"%red" = 100 - 0.4 = color(green)("99.6%")#

Because the pH of the solution is very low

$p {H}_{\text{solution}} = - \log \left(\left[{H}^{+}\right]\right) = - \log \left(0.05\right) = 1.3$

almost all of the methyl orange will be in acidic form.