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Chapter 4: Problem 75

What volume of \(0.0200 M\) calcium hydroxide is required to neutralize \(35.00 \mathrm{~mL}\) of \(0.0500 M\) nitric acid?

Short Answer

Expert verified
To neutralize 35.00 mL of 0.0500 M nitric acid, a volume of 43.75 mL of 0.0200 M calcium hydroxide is required.

Step by step solution

01

1. Write the balanced chemical equation

The balanced chemical equation for the reaction between calcium hydroxide and nitric acid is: \[Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O\] This equation tells us that one mole of calcium hydroxide reacts with two moles of nitric acid to produce one mole of calcium nitrate and two moles of water.
02

2. Calculate moles for nitric acid

We are given the volume and concentration of nitric acid, so to find the moles of nitric acid, we will use the following formula: Moles = volume × concentration Using the given values (volume = 35.0 mL and the concentration = 0.0500 M), we can calculate the moles of nitric acid: Moles of nitric acid = (35.0 mL) × (0.0500 mol/L) = 1.75 × 10^(-3) mol
03

3. Use the mole ratio to find moles of calcium hydroxide needed

From the balanced chemical equation, we know that 1 mol of calcium hydroxide reacts with 2 moles of nitric acid. So, we can find the moles of calcium hydroxide required using the mole ratio: Moles of calcium hydroxide = (moles of nitric acid) × (1 mol Ca(OH)_2 / 2 mol HNO_3) Moles of calcium hydroxide = (1.75 × 10^(-3) mol) × (1/2) = 8.75 × 10^(-4) mol
04

4. Calculate the volume of calcium hydroxide needed

We are given the concentration of the calcium hydroxide solution (0.0200 M), and we have calculated the moles of calcium hydroxide required (8.75 × 10^(-4) mol). To find the volume of calcium hydroxide needed, we use the formula: Volume = moles ÷ concentration Volume of Ca(OH)_2 = (8.75 × 10^(-4) mol) ÷ (0.0200 mol/L) = 0.04375 L Convert the volume to milliliters: Volume of Ca(OH)_2 = 0.04375 L × (1000 mL / 1 L) = 43.75 mL
05

5. Conclusion

To neutralize 35.00 mL of 0.0500 M nitric acid, a volume of 43.75 mL of 0.0200 M calcium hydroxide is required.

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Key Concepts

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

Neutralization Reaction
Neutralization reactions are a specific type of chemical reaction wherein an acid and a base react to form water and a salt. This process is rooted in the idea that acids donate protons (H+), while bases accept these protons. The strength and concentration of the reacting acid and base dictate the extent to which neutralization occurs.

In our exercise, we're dealing with a neutralization reaction between calcium hydroxide (Ca(OH)2), a strong base, and nitric acid (HNO3), a strong acid. According to the balanced chemical equation, \[Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O\],we observe that one mole of calcium hydroxide reacts with two moles of nitric acid to produce one mole of calcium nitrate and two moles of water. Recognizing these mole ratios is essential for stoichiometry calculations, which guide us in determining the quantities needed for complete neutralization.
Stoichiometry
Stoichiometry is the heart of chemical reaction calculations, providing the quantitative relationship between reactants and products in a chemical reaction. These calculations often involve mole ratios, which are derived from the coefficients of a balanced chemical equation.

In the example provided, the balanced equation informs us of the necessary mole ratio between calcium hydroxide and nitric acid.For every mole of calcium hydroxide, two moles of nitric acid are required. Thus, the stoichiometric calculation proceeds by first determining the moles of nitric acid using volume and molarity, then using the mole ratio to find the requisite amount of calcium hydroxide. With stoichiometry, we can convert these mole quantities into measurable volumes, allowing us to predict how much of a chemical is needed to react completely with a certain amount of another substance.
Molarity
Molarity, symbolized by 'M', is the measure of concentration of a solution in terms of moles of solute per liter of solution. It's a fundamental concept in chemistry, particularly when preparing solutions and carrying out titration calculations.

In this exercise, molarity tells us how many moles of each reactant are present in a given volume of solution. By providing the concentration of nitric acid (0.0500 M) and calcium hydroxide (0.0200 M), we have the information needed to calculate the volume of calcium hydroxide required to reach neutralization. By using the formula \[\text{Volume} = \frac{\text{moles}}{\text{concentration}}\],we can understand the inverse relationship between molarity and volume – as concentration increases, the required volume to achieve a specific amount of moles decreases, and vice versa. This precise evaluation of reactant quantities is crucial to achieving a successful neutralization reaction.

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Most popular questions from this chapter

A standard solution is prepared for the analysis of fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right)\), an anabolic steroid. A stock solution is first prepared by dissolving \(10.0 \mathrm{mg}\) of fluoxymesterone in enough water to give a total volume of \(500.0 \mathrm{~mL}\). A \(100.0-\mu \mathrm{L}\) aliquot (portion) of this solution is diluted to a final volume of \(100.0 \mathrm{~mL}\). Calculate the concentration of the final solution in terms of molarity.

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The VO \(^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \overline{\mathrm{SO}}_{4}\). A \(0.205-\mathrm{g}\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is \(0.298 \mathrm{~g}\). What mass of \(\mathrm{SO}_{4}{ }^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}{ }^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

When organic compounds containing sulfur are burned, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by reaction with hydrogen peroxide: $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ The resulting sulfuric acid is then titrated with a standard \(\mathrm{NaOH}\) solution. A \(1.325-\mathrm{g}\) sample of coal is burned and the \(\mathrm{SO}_{2}\) collected in a solution of hydrogen peroxide. It took \(28.44 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) to neutralize the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.

A solution is prepared by dissolving \(0.5842 \mathrm{~g}\) oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make \(100.0 \mathrm{~mL}\) of solution. A \(10.00-\mathrm{mL}\) aliquot (portion) of this solution is then diluted to a final volume of \(250.0\) \(\mathrm{mL}\). What is the final molarity of the oxalic acid solution?
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