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Chapter 4: Problem 45

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Short Answer

Expert verified
Answer: The percentage of sodium hydrogen carbonate in the sample is approximately \(71.61 \%\).

Step by step solution

01

Calculate moles of \(\mathrm{H}^{+}\) ions

To calculate the moles of \(\mathrm{H}^{+}\) ions, we will use the formula: Moles of \(\mathrm{H}^{+}\) = volume (in liters) × concentration (in moles per liter) Moles of \(\mathrm{H}^{+}\) = \(\frac{15.5 \: \mathrm{mL}}{1000}\) × \(0.275 \: \mathrm{M}\) = \(0.0042625 \: \mathrm{mol}\)
02

Calculate moles of \(\mathrm{NaHCO}_{3}\)

The balanced chemical equation given in the exercise states that one mole of \(\mathrm{H}^{+}\) reacts with one mole of \(\mathrm{NaHCO}_{3}\). Therefore, the moles of \(\mathrm{NaHCO}_{3}\) in the sample is equal to the moles of \(\mathrm{H}^{+}\) ions. Moles of \(\mathrm{NaHCO}_{3}\) = \(0.0042625 \: \mathrm{mol}\)
03

Calculate mass of \(\mathrm{NaHCO}_{3}\) in the sample

To calculate the mass of \(\mathrm{NaHCO}_{3}\), we will use the formula: Mass of \(\mathrm{NaHCO}_{3}\) = moles of \(\mathrm{NaHCO}_{3}\) × molar mass of \(\mathrm{NaHCO}_{3}\) Molar mass of \(\mathrm{NaHCO}_{3}\) = \(23 + 1 + 12 + 16 \times 3 = 84 g/mol\) Mass of \(\mathrm{NaHCO}_{3}\) = \(0.0042625 \: \mathrm{mol} \times 84 \: \mathrm{g/mol} = 0.35805 \: \mathrm{g}\)
04

Calculate the percentage of \(\mathrm{NaHCO}_{3}\) in the sample

The final step is to calculate the percentage of \(\mathrm{NaHCO}_{3}\) in the \(0.500 \: \mathrm{g}\) sample. To do this, we will use the formula: Percentage of \(\mathrm{NaHCO}_{3}\) = \(\frac{\text{Mass of } \mathrm{NaHCO}_{3}}{\text{Total mass of sample}} \times 100 \%\) Percentage of \(\mathrm{NaHCO}_{3}\) = \(\frac{0.35805 \: \mathrm{g}}{0.500 \: \mathrm{g}} \times 100 \% \approx 71.61 \%\) The percentage of sodium hydrogen carbonate in the sample is approximately \(71.61 \%\).

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Most popular questions from this chapter

Laws passed in some states define a drunk driver as one who drives with a blood alcohol level of \(0.10 \%\) by mass or higher. The level of alcohol can be determined by titrating blood plasma with potassium dichromate according to the unbalanced equation $$ \mathrm{H}^{+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$ Assuming that the only substance that reacts with dichromate in blood plasma is alcohol, is a person legally drunk if \(38.94 \mathrm{~mL}\) of \(0.0723 \mathrm{M}\) potassium dichromate is required to titrate a \(50.0\) -g sample of blood plasma?

Consider the following generic equation $$ \mathrm{OH}^{-}(a q)+\mathrm{HB}(a q) \longrightarrow \mathrm{B}^{-}(a q)+\mathrm{H}_{2} \mathrm{O} $$ For which of the following pairs would this be the correct prototype equation for the acid-base reaction in solution? If it is not correct, write the proper equation for the acid-base reaction between the pair. (a) hydrochloric acid and pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (b) sulfuric acid and rubidium hydroxide (c) potassium hydroxide and hydrofluoric acid (d) ammonia and hydriodic acid (e) strontium hydroxide and hydrocyanic acid

What volume of \(0.4163 M\) barium chloride will react completely with (a) \(12.45 \mathrm{~mL}\) of \(1.732 \mathrm{M}\) sulfuric acid? (b) \(15.00 \mathrm{~g}\) of ammonium phosphate? (c) \(35.15 \mathrm{~mL}\) of \(1.28 \mathrm{M}\) potassium carbonate?

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{N}_{2}(g)\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q)\) (c) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{4}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{IO}_{4}^{-}(a q) \longrightarrow \mathrm{IO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\)

Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus \(\left(P_{4}\right)\) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right)\) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.
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