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Chapter 4: Problem 137

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{7} .\) A \(0.250-\mathrm{g}\) sample of citric acid dissolved in \(25.0 \mathrm{~mL}\) of water requires \(37.2 \mathrm{~mL}\) of \(0.105 \mathrm{M}\) \(\mathrm{NaOH}\) for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

Short Answer

Expert verified
Citric acid has 3 acidic hydrogens per molecule, as determined through stoichiometry and mole calculations using the given mass, volume, and concentration values.

Step by step solution

01

Calculate the moles of NaOH used in the reaction

We know that the concentration of NaOH, c(NaOH), is 0.105 M, and the volume used to neutralize the citric acid is 37.2 mL. To determine the moles of NaOH used, we can use the following formula: n(NaOH) = c(NaOH) × V(NaOH) where n(NaOH) is the moles of NaOH, c(NaOH) is the concentration of NaOH, and V(NaOH) is the volume of NaOH in L. Remember to convert the given volume of NaOH from mL to L. n(NaOH) = 0.105 mol/L × (37.2 mL × 0.001 L/mL) n(NaOH) = 0.105 mol/L × 0.0372 L n(NaOH) = 0.003906 mol
02

Determine the moles of citric acid, H₃A

We know that the mass of the citric acid sample is 0.250 g. To find the moles of citric acid, we can use the molar mass of citric acid which is 192.124 g/mol (from the molecular formula C₆H₈O₇). Use the formula: n(H₃A) = mass(H₃A) / M(H₃A) where n(H₃A) is the moles of citric acid, mass(H₃A) is the mass of the citric acid sample, and M(H₃A) is the molar mass of citric acid. n(H₃A) = 0.250 g / 192.124 g/mol n(H₃A) = 0.00130 mol
03

Calculate the stoichiometric coefficient of hydrogens in the balanced equation

In the balanced equation, we have: H₃A + x NaOH -> NaₓA + 3x H₂O Let's find the stoichiometric coefficient (x) by dividing the number of moles of NaOH (0.003906 mol) by the number of moles of citric acid (0.00130 mol). x = n(NaOH) / n(H₃A) x = 0.003906 mol / 0.00130 mol x = 3 So, the balanced equation becomes: H₃A + 3 NaOH -> Na₃A + 3 H₂O
04

Determine the number of acidic hydrogens per molecule

The stoichiometric coefficient x we have calculated tells us that there are 3 moles of acidic hydrogens in each mole of citric acid. In other words, there are 3 acidic hydrogens per molecule of citric acid. Therefore, citric acid has 3 acidic hydrogens per molecule.

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Most popular questions from this chapter

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion b. calcium ion c. iron(III) ion d. sulfate ion e. mercury(I) ion, \(\mathrm{Hg}_{2}{ }^{2+}\) f. silver ion

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0\) ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals \(1.0 \mathrm{~g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q) \longrightarrow\) \(4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(t)\) This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

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