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## Q. 4.10

Many over-the-counter antacids, like the one shown in Figure 4.18, contain sodium bicarbonate. Sodium bicarbonate reacts with excess stomach acid (aqueous HCl) in a neutralization reaction. A 650 mg tablet is dissolved in 25.00 mL of water. Titration of this solution requires 25.27 mL of a 0.3000 M solution of HCl. (a) Write a net ionic equation for the reaction between $NaHCO_{3}$ and HCl. (b) What is the molarity of the antacid solution? (c) Is the tablet pure $NaHCO_{3}$?

## Verified Solution

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.