Question

An antacid tablet (such as Tums or Rolaids) weighs $\mathbf{1}\mathbf{.}\mathbf{3259}\mathbf{G}$. The only acid-neutralizing ingredient in this brand of antacid is ${\mathbf{CaCO}}_{\mathbf{3}}$. When placed in $\mathbf{12}\mathbf{.}\mathbf{07}\mathit{m}\mathit{L}$of $\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{}\mathit{M}\mathit{}\mathbf{HCl}$, the tablet fizzes merrily as ${\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ is given off. After all of the ${\mathbf{CO}}_{\mathbf{2}}$ has left the solution, an indicator is added, followed by $\mathbf{11}\mathbf{.}\mathbf{74}\mathit{m}\mathit{L}\mathit{}$of $\mathbf{0}\mathbf{.}\mathbf{531}\mathbf{M}\mathbf{}\mathbf{NaOH}$. The indicator shows that at this point the solution is definitely basic. The addition of$\mathbf{5}\mathbf{.}\mathbf{12}\mathit{m}\mathit{L}$ of $\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{M}\mathbf{}\mathbf{HCl}$makes the solution acidic again. Then $\mathbf{3}\mathbf{.}\mathbf{17}\mathit{m}\mathit{L}$ of the $\mathbf{0}\mathbf{.}\mathbf{513}\mathbf{M}\mathbf{}\mathbf{NaOH}$ brings the titration exactly to an endpoint, as signaled by the indicator. Compute the percentage by mass of ${\mathbf{CaCO}}_{\mathbf{3}}$in the tablet.

Short answer

The percentage by mass of calcium carbonate, ${{\mathrm{CaCO}}_3}_{}$in the antacid tablet is $39.5\%$.

Step-by-step solution

Given Steps of Chemical Reactions.

The following are the steps in the antacid tablet analysis:

Step-1: $12.07\mathrm{mL}\mathrm{of}1.07\mathrm{M}\mathrm{HCl}$is added to $1.3259\mathrm{g}$of antacid tablet.

The reaction between calcium carbonate $\left({\mathrm{CaCO}}_3\right)$ in the tablet and acid ${\mathrm{H}}_3{\mathrm{O}}^{+}$ is as follows:

${\mathrm{CaCO}}_3\left(s\right)+2{\mathrm{h}}_3{\mathrm{o}}^{+}\left(aq\right)\to {\mathrm{Ca}}^{2+}\left(aq\right)+{\mathrm{CO}}_2\left(g\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

The mole ratio of calcium carbonate to acid is $1:2$.

Step-2: Addition of $11.74\mathrm{mL}\mathrm{of}0.531\mathrm{M}\mathrm{NaOH}$to the solution mixture of step-1 makes the solution basic. So, the acid remains in the solution after step-1 is completely neutralized.

The reaction between sodium hydroxide $\left(\mathrm{NaOH}\right)$and acid $\left({\mathrm{H}}_3{\mathrm{O}}^{+}\right)$is as follows:

$\mathrm{NaOH}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\to {\mathrm{Na}}^{+}\left(aq\right)+2{\mathrm{H}}_2\mathrm{O}\left(l\right)$

The mole ratio of sodium hydroxide to add is $1:1$.

Step-3: The addition of $5.12\mathrm{mL}\mathrm{of}1.07\mathrm{M}\mathrm{HCl}$to the solution of the mixture after step-2 makes the solution acidic. So, the base remains in the solution after step-2 is completely neutralized.

The reaction between sodium hydroxide $\left(\mathrm{NaOH}\right)$ and acid $\left({\mathrm{H}}_3{\mathrm{O}}^{+}\right)$ is as follows:

$\mathrm{NaOH}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\to {\mathrm{Na}}^{+}\left(aq\right)+2{\mathrm{H}}_2\mathrm{O}\left(l\right)$

The mole ratio of sodium hydroxide to add is $1:1$.

Step-4: Addition of ofto the solution of mixtures after step-3 makes the solution neutral.

The reaction between sodium hydroxide $\left(\mathrm{NaOH}\right)$ and acid $\left({\mathrm{H}}_3{\mathrm{O}}^{+}\right)$ is as follows:

$\mathrm{NaOH}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\to {\mathrm{Na}}^{+}\left(aq\right)+2{\mathrm{H}}_2\mathrm{O}\left(l\right)$

The mole ratio of sodium hydroxide to add is $1:1$.

Finding No. of Moles of NaOH

Calculate the total moles of acid added in step-1 and step-3 and calculate the total moles of sodium hydroxide added in step-2 and step-4. Subtract the total moles of sodium hydroxide from the total moles of acid. This gives the moles of acid to react with calcium carbonate.

Multiply the volume and molarity to get the number of moles. Thus, the total moles of $\mathrm{HCl}$ is:

$$\begin{gathered} \mathrm{Total}\mathrm{moles}\mathrm{of}\mathrm{HCl}=\left(12.07\mathrm{mL}\times 1.07\mathrm{M}\right)+\left(5.12\mathrm{mL}\times 1.07\mathrm{M}\right) \\ =12.9149+5.4784 \\ =18.39\mathrm{mmol} \end{gathered}$$

The total moles of $\mathrm{NaOH}$is:

$$\begin{gathered} \mathrm{Total}\mathrm{moles}\mathrm{of}\mathrm{NaOH}=\left(11.74\mathrm{mL}\times 0.531\mathrm{M}\right)+\left(3.17\mathrm{mL}\times 0.531\mathrm{M}\right) \\ =6.2339+1.6832 \\ =7.92\mathrm{mmol} \end{gathered}$$

Subtract the total moles of sodium hydroxide from the total moles of acid:

$18.39-7.92=10.47\mathrm{mmol}$

Therefore, the number of moles of acid that reacts with calcium carbonate is $10.47\mathrm{mmol}$.

Calculating mass of CaCO3.

Calculate the number of moles of calcium carbonate in the table by multiplying the number of moles of acid that react with calcium carbonate with the mole ratio of calcium carbonate to acid.

$$\begin{gathered} \mathrm{mol}{\mathrm{CaCO}}_3=10.47\mathrm{mmol}\mathrm{acid}\times \frac{1\mathrm{mol}}{1000\mathrm{mmol}}\times \frac{1\mathrm{mol}{\mathrm{CaCO}}_3}{2\mathrm{mol}\mathrm{acid}} \\ =5.235\times {10}^{‐3} \end{gathered}$$

Calculate the mass of calcium carbonate by multiplying its moles with its molar mass $\left(100.09\mathrm{g}/\mathrm{mol}\right)$.

$5.235\times {10}^{‐3}\mathrm{mol}{\mathrm{CaCO}}_3\times \frac{100.09\mathrm{g}}{1\mathrm{mol}}=0.524\mathrm{g}{\mathrm{CaCO}}_3$

Calculate the percent by mass of calcium carbonate by dividing mass of calcium carbonate by mass of tablet and then multiply with 100.

$\frac{0.524\mathrm{g}}{1.3259\mathrm{g}}\times 100=39.5\%$

Hence, the percentage by mass of calcium carbonate in the antacid tablet is $39.5\%$.