Answer to Question #141922 in Chemistry for Roaa Mohammed

Question

50.00 ml of nitric acid (of unknown
concentration) was pipetted into a
150 ml flask and an indicator was
added. 0.1 mol/L NaOH (a strong
base) was added gradually to the
flask from a buret.
a) Write a symbolic equation for
this chemical reaction.
b) Explain at the molecular level
why the pH increases as NaOH is
added to HNO3.
c) What is the volume of NaOH
at the equivalence point?
d) What is the pH at the
equivalence point?
e) Calculate the concentration of
the nitric acid solution.

Answer 1

A symbolic equation for the chemical reaction between nitric acid and sodium hydroxide is:

HNO₃ + NaOH $\rightarrow$ NaNO₃ + H₂O.

As HNO₃ and NaOH are strong acid and strong base, respectively, they dissociate almost completely in water solution and form ions. Therefore, the reaction equation can be rewritten as:

H⁺ + OH^- $\rightarrow$ H₂O.

From this equation, one can see that upon addition of NaOH, protons H⁺ react with OH^- and form water H₂O. Therefore, the concentration $[H^+]$ decreases both because of dilution and the decrease of H⁺ amount in the solution. According to its definition, $pH = -\text{log}[H^+]$ . Thus, pH increases as NaOH is added to HNO₃.

At the equivalence point, chemically equivalent quantities of reactants have been mixed. This means that the number of the moles of HNO₃ initially present in the solution and the number of the moles of NaOH added are equal:

$n(HNO_3) = n(NaOH)$ .

The solution at the equivalence point contains salt NaCl and water. As both acid and base are strong, the solution at the equivalence point is neutral, so that the pH at the equivalence point is 7.00. The volume of NaOH at the equivalence point is the volume added to reach the pH=7.00.

Finally, the concentration of the nitric acid solution is calculated from the relation shown above and the volume of NaOH needed to reach the pH=7.00:

$c(HNO_3)=\frac{V_{NaOH, pH=7.00}\cdot c_{NaOH}}{V(HNO_3)} = \frac{V_{NaOH, pH=7.00}\cdot 0.1\text{mol/L}}{50.00 \text{mL}}$ mol/L.