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The active component in one type of calcium dietary supplement is calcium carbonate. A \(1.2450 \mathrm{g}\) tablet of the supplement is added to \(50.00 \mathrm{mL}\) of \(0.5000 \mathrm{M} \mathrm{HCl}\) and allowed to react. After completion of the reaction, the excess HCl(aq) requires \(40.20 \mathrm{mL}\) of \(0.2184 \mathrm{M}\) NaOH for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of \(\mathrm{Ca}^{2+} ?\)

Short Answer

Expert verified
The calcium content of the tablet is 325 mg.

Step by step solution

01

Identify Reactions

Identify the reactions involved in the problem. The first reaction is the reaction of the calcium carbonate (CaCO3) in the supplement with hydrochloric acid (HCl). This reaction can be expressed as follows: \(CaCO3 + 2HCl \rightarrow CaCl2 + CO2 + H2O\). The second reaction is the titration of the residual HCl with sodium hydroxide (NaOH), which can be expressed as: \(HCl + NaOH \rightarrow NaCl + H2O\).
02

Calculate Moles of HCl in Excess

Calculate the moles of the HCl that react with NaOH. The amount of a substance in moles can be calculated by multiplying the volume by the molarity. Thus, the moles of HCl in excess is \(40.20 \mathrm{mL} \times 0.2184 \mathrm{M} = 0.00877 \mathrm{mol}\).
03

Determine Moles of HCl Reacted with Tablet

As the total amount of HCl is known, the amount of HCl that reacted with the tablet can be calculated by subtracting the excess from the total amount. Therefore, \(0.02500 \mathrm{mol} - 0.00877 \mathrm{mol} = 0.01623 \mathrm{mol}\). This is the amount of HCl that reacted with the calcium carbonate in the tablet.
04

Determine Moles of CaCO3

From the balanced chemical equation for the reaction between CaCO3 and HCl, it is seen that one mole of CaCO3 react with two moles of HCl. Therefore, the moles of CaCO3 in the tablet is \(0.01623 \mathrm{mol} / 2 = 0.00811 \mathrm{mol}\).
05

Calculate Mass of Calcium

The mass of calcium in the tablet can be determined by multiplying the moles of CaCO3 by the molar mass of calcium. Therefore, the mass of calcium in the tablet is \(0.00811 \mathrm{mol} \times 40.08 \mathrm{g/mol = 0.325 \mathrm{g}\). This value needs to be converted to milligrams by multiplying by 1000, which yields 325 mg of calcium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calcium Content Determination
Calcium content determination in supplements is crucial for understanding the precise amount of calcium supplied to the body. In our scenario involving a calcium dietary supplement, the core compound is calcium carbonate (CaCO₃). By measuring the calcium ion content in a tablet, we can assess its efficiency and potency.

The process begins with a chemical reaction where the CaCO₃ in the tablet reacts with hydrochloric acid (HCl). This releases the calcium ions into a solution. The reaction transforms CaCO₃ into calcium chloride (CaCl₂), carbon dioxide (CO₂), and water (H₂O). The goal is to identify the calcium available post-reaction to determine the actual content in the supplement.

Performing an excess and back titration method using sodium hydroxide (NaOH) allows one to calculate the remaining HCl. This reveals how much HCl initially reacted with CaCO₃, indicating the amount of calcium present.
Chemical Titration
Chemical titration is a laboratory technique used to determine the concentration of a solute in a solution. It is central to our exercise where we establish how much hydrochloric acid reacted with the calcium carbonate tablet. In this context, back titration is used, which measures excess reactant left after the primary reaction.
  • Back Titration Method: Initially, an excess amount of HCl is reacted with the calcium carbonate tablet.
  • The residual HCl after the reaction is then titrated using a known concentration of NaOH.
This method is helpful when the end point of a direct titration is not easily detected or when slow reactions are involved.

By knowing the precise volume and concentration of NaOH used to neutralize this excess HCl, one can back-calculate the amount of HCl that initially reacted with the tablet. Such precision is fundamental in determining the exact quantity of calcium in calcium carbonate supplements.
Stoichiometry
Stoichiometry plays a crucial role in understanding chemical quantities and ensuring balance in reactions. In our example, it helps decipher the exact proportions required for the reaction between calcium carbonate and hydrochloric acid.

The balanced equation for this reaction is:\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]This equation shows that one mole of calcium carbonate reacts with two moles of hydrochloric acid. By applying stoichiometry principles, we can calculate the number of moles of each reactant used and the products formed.
  • Mole Ratio: Knowing the molar ratio (1:2) helps determine how many moles of HCl are needed for a given amount of CaCO₃.
  • Conversion: Utilize molar masses to convert between grams and moles, crucial for determining the mass of calcium from its moles.
Mastering stoichiometry is vital for any successful chemistry experiment, ensuring reactions are quantitatively accurate.

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Most popular questions from this chapter

When treated with dilute \(\mathrm{HCl}(\mathrm{aq}),\) the solid that reacts to produce a gas is (a) \(\mathrm{BaSO}_{3} ;\) (b) \(\mathrm{ZnO};\) (c) \(\mathrm{NaBr} ;\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) \(\mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow\) (b) \(\mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow\) (c) \(\operatorname{FeS}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow\) (d) \(\mathrm{K}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow\) (e) \(\mathrm{Mg}(\mathrm{s})+\mathrm{H}^{+} \longrightarrow\)

How many milligrams of \(\mathrm{MgI}_{2}\) must be added to \(250.0 \mathrm{mL}\) of \(0.0876 \mathrm{M} \mathrm{KI}\) to produce a solution with \(\left[\mathrm{I}^{-}\right]=0.1000 \mathrm{M} ?\)

The active ingredients in a particular antacid tablet are aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2} . \quad \mathrm{A} 5.00 \times 10^{2} \mathrm{mg}\) sample of the active ingredients was dissolved in \(50.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl} .\) The resulting solution, which was still acidic, required \(16.5 \mathrm{mL}\) of \(0.377 \mathrm{M} \mathrm{NaOH}\) for neutralization. What are the mass percentages of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the sample?

In the equation \(\begin{aligned} ? \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq}) & \longrightarrow ? \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \end{aligned}\) the missing coefficients (a) are each \(2 ;\) (b) are each 4; (c) can have any values as long as they are the same; (d) must be determined by experiment.

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