(a) What is the pH at the equivalence point when 0.030 0 M NaF is titrated with 0.060 0 M HClO4?

(b) Why would an indicator endpoint probably not be useful in this titration?

An acid-base titration is a titration between acid and base. It is also known as a neutralization reaction.

There are primarily four types of acid-base titrations -

• Strong base vs strong acid

• Strong base vs weak acid

• Weak base vs strong acid

• Weak base vs weak acid

The equivalence point ensures the titration reaction becomes completed and at this point, the number of moles of titrant and the number of moles of analyte remains equal. We cannot see with naked eyes the completion of the titration reaction - the titrant and the analyte becoming equal in stoichiometric proportion, hence we use a chemical compound called indicator called which indicates the end of the reaction by color change. The endpoint refers to the color change and indicates the equivalence point has been reached.

_{$\mathrm{NaF}$} completely dissociates into role="math" localid="1663552868742" ${\mathrm{Na}}^1$and ${\mathrm{F}}^{-}$ions in the aqueous medium. The ${\mathrm{F}}^{-}$ions grab a proton from perchloric acid$\left({\mathrm{HClO}}_4\right)$ and forms $\mathrm{HF}.\mathrm{HF}$dissociates into $\mathrm{H}$and $\mathrm{F}$ions. So the actual reaction that occurs when $\mathrm{NaF}$ is titrated with ${\mathrm{HClO}}_4$ is the dissociation of $\mathrm{HF}$ and the $\mathrm{pH}$ is determined the concentration of$\mathrm{HF}$and ${\mathrm{HClO}}_4$ present during the course of titration.

Given data:

strength of $\mathrm{NaF}=0.0300\mathrm{M}$

strength of ${\mathrm{HClO}}_4=0.0600\mathrm{M}$

Concentration of ${\mathrm{HClO}}_4$ is two times greater than that of$\mathrm{NaF}$that if$\mathrm{VmL}$of$\mathrm{NaF}$is used half of the volume of ${\mathrm{HClO}}_4\mathrm{NaF}$is the volume of${\mathrm{HClO}}_4$required to reach the equivalence point. One mole of$\mathrm{NaF}$completely dissociates into one mole of HF that the volume of NaF taken completely converted to HF. Thus formal concentration of HF at the equivalence point is calculated as,

role="math" localid="1663554102969" $\left(\frac{\mathrm{V}}{\mathrm{V}+{\displaystyle \frac{1}{2}}\mathrm{v}}\right)\times 0.0300\mathrm{M}=0.0200\mathrm{M}$

The pH is determined by dissociation of HF as,

$\mathrm{HF}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{F}}^{-}$

Formal concentration of HF is calculated to be 0.0200 M and so, $\left[\mathrm{HF}\right]=0.0200\mathrm{M},\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{F}}^{-}\right]=\mathrm{x}.$ already known that Ka for dissociation of HF is $3.36\times {10}^{-3}$and so,

$\mathrm{Ka}=\frac{{\mathrm{x}}^2}{(0.0200-\mathrm{x})}$

role="math" localid="1663554671633" $6.6\times {10}^4=\frac{{\mathrm{x}}^2}{0.0200-\mathrm{x}}$

Solving for ' $\mathrm{X}$',

$\mathrm{x}=3.36\times {10}^{-3}$

Therefore,

$\left[{\mathrm{H}}^{+}\right]=3.36\times {10}^{-3}$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =2.47 \end{gathered}$$

As calculated in the previous step equivalence point occurs at pH 2.47 which is quite low that the inflection in the titration curve is not observed. Since there is no drastic change is observed in pH the notable change in color of the indicator which is influenced by pH change, is not observed. Hence the usage of indicator for this titration is probably prove to be not useful.