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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

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

On the basis of the general solubility rules given in Table 4.1, predict which of the following substances are likely to be soluble in water. a. aluminum nitrate b. magnesium chloride c. rubidium sulfate d. nickel(II) hydroxide e. lead(II) sulfide f. magnesium hydroxide g. iron(III) phosphate

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 \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) $$ 4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ 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?

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} .\) A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of \(\mathrm{NaOH}\) is added. producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3} .\) The precipitate is filtered. dried, and weighed. Its mass is \(0.107 \mathrm{~g}\). Calculate the following. a. the mass of iron in the sample b. the mass of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample.,
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