# Question aa498

Aug 12, 2015

Here's how you can approach this problem.

#### Explanation:

FULL QUESTION

M–nitrophenol, a weak acid, can be used as an indicator because its base, "In"^(–), colour is yellow and its acid, $\text{HIn}$, is colourless.

The pH of a 0.010M solution of the compound is 3.44. What would be the lowest pH at which a solution of the indicator would have a definite yellow colour?

I'll start by saying that your values could be a little off, because this concentration of m-nitrophenol would actually produce a pH of about 5.1 to 5.3.

I'll show you the concept behind this problem and you can double-check your values if you want.

So, what this problem wants you to do is recognize the fact that you're dealing with a weak acid that does not dissociate completely in aqueous solution.

When placed in aqueous solution, this acid will exist both as an acid, and a conjugate base, the ratio beyween these two forms ultimately being determined by the solution's pH.

The degree of dissociation for any weak acid depends on its acid dissociation constant, ${K}_{a}$. The equilibrium reaction that is established when m-nitrophenol reacts with water looks like this

$H I {n}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + I {n}_{\left(a q\right)}^{-}$

The first thing that you need to do is determine the value of the acid dissociation constant. To do that, use the fact that the concentration of hydronium ions can be calculated using pH by

$\left[{H}_{3} {O}^{+}\right] = {10}^{- p H}$

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 3.44} = 3.63 \cdot {10}^{- 4} \text{M}$

You can work your way backward from this value to determine ${K}_{a}$ by using an ICE table

$H I {n}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + I {n}_{\left(a q\right)}^{-}$

I......0.010....................................0..................0
C.......(-x).....................................(+x)...............(+x)
E.....0.010-x.................................x...................x

You know that $x$, which represetns the equilibrium concentration of both the hydronium ions, and the conjugate base ${\text{In}}^{-}$, is equal to $3.63 \cdot {10}^{- 4}$. This means that ${K}_{a}$ is

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[I {n}^{-}\right]}{\left[H I n\right]}$

${K}_{a} = \frac{3.63 \cdot {10}^{- 4} \cdot 3.63 \cdot {10}^{- 4}}{0.010 - 3.63 \cdot {10}^{- 4}} = 1.37 \cdot {10}^{- 5}$

Now, a very useful tool to have when dealing with an acid is its $p {K}_{a}$, which is equal to

$p {K}_{a} = - \log \left({K}_{a}\right)$

$p {K}_{a} = - \log \left(1.37 \cdot {10}^{- 5}\right) = 4.86$

The value of the $p {K}_{a}$ will tell you at what pH will the molecule go from behaving as an acid, i.e. donating a proton, to behaving as a base, i.e. accepting a proton.

Since the acid's dissociation is an equilibrium reaction, think of what will happen if you decrease pH.

When you decrease pH, you actually increase the concentration of hydronium ions (or protons, ${H}^{+}$), which means tha the equilibrium will shift to the left.

The concentration of the conjugate base will decrease as a result of this shift. The weak acid will be the dominant form and the solution will be colorless. This happens when pH is smaller than $p {K}_{a}$.

On the other hand, is you increase the pH, i.e. decrease the concentration of the hydronium ions, the equilibrium will shift to the right.

More acid will be consumed and more conjugate base will be produced. If the pH is high enough, then the conjugate base will be the dominant form and the solution will be yellow.

At $p {H}_{\text{sol}} = p {K}_{a}$, both species are present in equal amounts.

So, in order for the solution to have a definite yellow color, you need the conjugate base form to dominate the solution. This means that the pH must be greater than $p {K}_{a}$.

Now, the color change for m-nitrophenol takes place over a pH range of about $p {K}_{a} \pm 1$, which means that at

$p {H}_{\text{sol}} = 4.86 + 1 = \textcolor{g r e e n}{5.86}$

the solution would have a definite yellow color.

SIDE NOTE You could also use the Henderson-Hasselbalch equation to try and explain why this is the case

pH_"sol" = pK_a + log((["conjugate base"])/(["weak acid"]))#

Once again, a pH smaller than $p {K}_{a}$ implies that the log is negative, hence you have more acid than conjugate base.

When pH is bigger than $p {K}_{a}$, the log is positive, hence you have more conjugate base than weak acid.