Question

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 mL of the weak acid solution has been added to 50.0 mL of the 0.100 M NaOH solution, the pH of the resulting solution is 10.50. Calculate the original concentration of the solution of weak acid.

Short answer

The original concentration of monoprotic acid is0.210 M.

Step-by-step solution

The concentration of [OH−].

The reaction of weak monoprotic acid can be denoted as,

$HA\left(aq\right)+NaOH\left(aq\right)\to NaA\left(aq\right)+H_{2}O\left(l\right)$

The pH of the solution is 10.50. Then, as

$pH+pOH=14$

So here,

$\begin{array}{c}pOH=14-10.50\\ =3.5\end{array}$

Also,

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ 3.5=-\log \left[O{H}^{-}\right]\end{array}$

Hence,

$\begin{array}{l}\left[O{H}^{-}\right]={10}^{-3.5}\\ \left[O{H}^{-}\right]=3.16\times {10}^{-4}M\end{array}$

The concentration of monoprotic acid.

The number of moles of remaining are the multiplication of molarity and volume.

As,

$\begin{array}{l}Molarity=\frac{moles}{Volume}\\ So,\\ moles=M\times V\end{array}$

Hence,

$\begin{array}{l}=\frac{3.16\times {10}^{-4}\times 73.75}{1000}\\ =2.33\times {10}^{-5}moles\end{array}$

Also, the initial molarity and volume taken can be 0.1 M and 50 ml.

$\begin{array}{c}TheinitialmolesofO{H}^{-}=M_{NaOH}\times V_{NaOH}\\ =\frac{0.1\times 50}{1000}\\ =5\times {10}^{-3}moles\end{array}$

The number of moles in original concentration of the solution of weak acid is

$\begin{array}{c}n=5\times {10}^{-3}-2.33\times {10}^{-5}\\ =4.97\times {10}^{-3}moles\end{array}$

Then, theoriginal concentration of the solution of the weak acid is

$\begin{array}{c}Concentrationofacid=\frac{numberofmoles}{Volumeinliter}\\ =\frac{4.97\times {10}^{-3}\times 1000}{23.75}\\ =0.2092M\\ =0.21M\end{array}$