Collect and Organize We are assigned three tasks in this sample exercise. We are given the names of the reactants and asked to write a balanced net ionic equation. We are given the volume and concentration of the titrant and the volume of the sample and asked to calculate the concentration of the solution. Finally, using the results of our calculations, we are asked to assess the purity of the NaHCO_{3} present in the original sample.

Analyze (a) The products of the reaction between NaHCO_{3} and HCl are sodium chloride, water, and carbon dioxide. We will first balance the molecular equation and then generate the overall ionic equation and the net ionic equation. (b) The steps involved in calculating the molarity of the sodium bicarbonate solution (not including liter–milliliter conversions) are

(c) The purity of the tablet can be determined by converting the number of moles of NaHCO_{3} calculated in part (b) into an equivalent number of grams:

and comparing the result with the mass of the tablet (650 mg).

Solve
a.  The balanced molecular equation for the neutralization reaction is

NaHCO_{3}(aq) + HCl(aq) → H_{2}O(\ell) + CO_{2}(g) + NaCl(aq)

The balanced ionic equation is

Na^{+}(aq) + HCO_{3}^{-} (aq) + H^{+}(aq) + Cl^{-}(aq) → H_{2}O(\ell) + CO_{2}(g) + Na^{+}(aq) + Cl^{-}(aq)

and the net ionic equation is

\sout{Na^{+}(aq)} + HCO_{3}^{-} (aq) + H^{+}(aq) + \sout{Cl^{-}(aq)} → H_{2}O(\ell) + CO_{2}(g) + \sout{Na^{+}(aq)} + \sout{Cl^{-}(aq)}

HCO_{3}^{-} (aq) + H^{+}(aq) → H_{2}O(\ell) + CO_{2}(g)

b.  The stoichiometry of the reaction shows that 1 mole of NaHCO_{3} reacts with 1 mole of HCl. The number of moles of NaHCO_{3} titrated is

0.02527  \sout{L  HCl} \times \frac{0.3000  \sout{mol  HCl}}{1  \sout{L  HCl}} \times \frac{1  mol  NaHCO_{3}}{1  \sout{mol  HCl}}=0.007581  mol  NaHCO_{3}

The molarity of the HCO_{3}^{-} is

\frac{0.007581  mol  NaHCO_{3}}{0.02500  L  solution}=0.3032  M  NaHCO_{3}

c.  To calculate the number of grams of sodium bicarbonate in the sample, we first need to know the molar mass of NaHCO_{3}:

\mathscr{M}_{NaHCO_{3}}=22.99  g/mol + 1.008  g/mol + 12.01  g/mol + 3(16.00  g/mol)

=84.01  g/mol  NaHCO_{3}

Therefore, the mass of the tablet is

0.007581  \sout{mol  NaHCO_{3}} \times \frac{84.01  g  NaHCO_{3}}{1  \sout{mol  NaHCO_{3}}}=0.6369  g  NaHCO_{3}

The mass of the tablet was 650 mg. On the basis of the titration, the sample contained only 637 mg of NaHCO_{3} and hence is not pure NaHCO_{3}.

Think About It The stoichiometry of the net ionic equation indicates a 1:1 ratio of acid to base. Because a nearly equivalent volume of aqueous HCl was needed to titrate 25 mL of antacid solution, it makes sense that the concentration of the base is similar to that of the acid.