Question

A 10.00-g sample of the ionic compound NaA, where A is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 500.0 mL HCl was added, the pH was 5.00. The experimenter found that 1.00 L of 0.100 M HCl was required to reach the stoichiometric point of the titration.

a.What is the molar mass of NaA?

b.Calculate the pH of the solution at the stoichiometric point of the titration.

Short answer

The molar mass of NaA is100 g/mol. The pH of the solution at the stoichiometric point is 3.02

Step-by-step solution

Subpart (a) The molar mass of NaA.

The chemical equation can be written as,

$NaA+HCl\to NaCl+HA$

There is a requirement of 1 mole of HCl for titration of 1 mole of NaA.

Thus,10 g of NaA has 0.1 mole of NaA.

$\begin{array}{c}molesofNaA=\left(1L\right)\frac{\left(0.1mol\right)}{L}\\ =0.1molofNaA\end{array}$

Therefore,

$\begin{array}{c}molarmassofNaA=\frac{massofNaA}{molesofNaA}\\ =\frac{10g}{0.1mol}\\ =100g/mol\end{array}$

Subpart(b) The pH of the solution at the stoichiometric point of the titration.

At the half-equivalence point,

$\begin{array}{l}pH=p{K}_a+{\log }_{10}1\\ pH=p{K}_a+0\\ p{K}_a=5\end{array}$

Also,

$\begin{array}{l}{K}_a={10}^{-5}\\ K_a=1\times {10}^{-5}\end{array}$

Here, all ${\text{A}}^{-}$converted to HA at the stoichiometric point.

$\begin{array}{c}\left[HA\right]=\frac{molesofHA}{totalvolume}\\ =\frac{0.1mol}{1L+0.1L}\\ =0.0909M\end{array}$

Now,

$\begin{array}{c}{K}_a=\left[HA\right]\\ K_a=\frac{x^2}{\left[HA\right]-x}\\ 1\times {10}^{-5}=\frac{x^2}{0.0909-x}\end{array}$

Therefore,

$x=9.48\times {10}^{-4}M$

$x=\left[H^{+}\right]=9.48\times {10}^{-4}M$

And,

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pH=-\log (9.48\times {10}^{-4}M)\\ pH=3.02\end{array}$