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Chapter 5: Problem 86

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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Most popular questions from this chapter

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

Household ammonia, used as a window cleaner and for other cleaning purposes, is \(\mathrm{NH}_{3}(\mathrm{aq}) .\) The \(\mathrm{NH}_{3}\) present in a 5.00 \(\mathrm{mL}\) sample is neutralized by 28.72 \(\mathrm{mL}\) of \(1.021 \mathrm{M} \mathrm{HCl}\). The net ionic equation for the neutralization is \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})\) What is the molarity of \(\mathrm{NH}_{3}\) in the sample?

An unknown white solid consists of two compounds, each containing a different cation. As suggested in the illustration, the unknown is partially soluble in water. The solution is treated with \(\mathrm{NaOH}(\mathrm{aq})\) and yields a white precipitate. The part of the original solid that is insoluble in water dissolves in \(\mathrm{HCl}(\mathrm{aq})\) with the evolution of a gas. The resulting solution is then treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and yields a white precipitate. (a) Is it possible that any of the cations \(M g^{2+}, C u^{2+}\) \(\mathrm{Ba}^{2+}, \mathrm{Na}^{+},\) or \(\mathrm{NH}_{4}^{+}\) were present in the original unknown? Explain your reasoning. (b) What compounds could be in the unknown mixture (that is, what anions might be present)?

Assign oxidation states to the elements involved in the following reactions. Indicate which are redox reactions and which are not. (a) $\mathrm{MgCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow$ $\mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})$ (b) $\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})$ (c) $\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow$ $\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}_{2}(\mathrm{g})$ (d) $2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})$
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