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Chapter 4: Problem 69

What volume of each of the following acids will react completely with \(50.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\) ? a. \(0.100 \mathrm{M} \mathrm{HCl}\) b. \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) ( 1 acidic hydrogen)

Short Answer

Expert verified
The volume of each acid needed to react completely with 50.00 mL of 0.200 M NaOH is: a. 100.0 mL of 0.100 M HCl b. 66.7 mL of 0.150 M HNO3 c. 50.0 mL of 0.200 M HC2H3O2

Step by step solution

01

Write the balanced chemical equation for the reaction.

For each of the acids reacting with NaOH, the balanced chemical equations are: a. \(HCl + NaOH \rightarrow NaCl + H_2O\) b. \(HNO_3 + NaOH \rightarrow NaNO_3 + H_2O\) c. \(HC_2H_3O_2 + NaOH \rightarrow NaC_2H_3O_2 + H_2O\) Notice that in each case, the stoichiometric ratio between the acid and NaOH is 1:1.
02

Calculate the moles of NaOH.

First, we need to find the moles of NaOH, given its volume and molarity: Moles of NaOH = Molarity × Volume Moles of NaOH = 0.200 mol/L × 50.00 mL × (1 L / 1000 mL) = 0.0100 mol.
03

Use stoichiometry to find moles of each acid needed to react completely.

Since the ratio of the acid to NaOH is 1:1, the moles of each acid needed to react completely with 0.0100 mol NaOH will be the same: a. Moles of HCl = 0.0100 mol b. Moles of HNO3 = 0.0100 mol c. Moles of HC2H3O2 = 0.0100 mol
04

Calculate the volume of each acid required.

Now we can use the concentration of each acid to find the volume needed to provide 0.0100 mol: a. Volume of HCl = moles / concentration = 0.0100 mol / 0.100 mol/L = 0.100 L or 100.0 mL b. Volume of HNO3 = moles / concentration = 0.0100 mol / 0.150 mol/L = 0.0667 L or 66.7 mL c. Volume of HC2H3O2 = moles / concentration = 0.0100 mol / 0.200 mol/L = 0.0500 L or 50.0 mL So, the volume of each acid needed to react completely with 50.00 mL of 0.200 M NaOH is: a. 100.0 mL of 0.100 M HCl b. 66.7 mL of 0.150 M HNO3 c. 50.0 mL of 0.200 M HC2H3O2

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Key Concepts

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

Stoichiometry
Stoichiometry is the cornerstone of chemical reactions where it involves the calculation of reactants and products in chemical processes. This field of chemistry ensures that equations are balanced and that the relationship between the quantity of reactants and products follows the law of conservation of mass.

Understanding the concept is crucial when solving acid-base titration problems such as the one in our exercise. Here, we work with a 1:1 stoichiometric ratio, meaning for every one mole of acid, one mole of NaOH is needed for the reaction to occur. This relationship is foundational and must be grasped to predict the outcomes of these reactions efficiently.
Molarity
Molarity, abbreviated as 'M', is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute divided by the volume of the solution in liters.

When solving titration exercises, understanding molarity is essential since it helps determine how much of a substance is present in a certain volume of solution. The exercise provided uses molarity to establish how much acid reacts with a given volume and concentration of NaOH solution. Being comfortable with this concept allows students to solve such problems with ease.
Balanced Chemical Equations
Balanced chemical equations are pivotal as they illustrate the law of conservation of mass, ensuring that atoms are neither created nor destroyed in a chemical reaction. Each equation lists the reactants on the left, an arrow representing the reaction, and the products on the right.

In the context of our titration exercise, the equations show the reactants (the acid and NaOH) and the products (salt and water). They are balanced with the same number of atoms on each side, reflecting the stoichiometric ratio which informs the calculation of reactants' and products' quantities essential for the titration process.
Moles Calculation
The concept of moles is central to chemistry; it is the bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. In simple terms, a mole corresponds to Avogadro's number, which is approximately 6.02 x 10^23 particles.

In the given exercise, moles calculation allows us to convert the volume and molarity of a NaOH solution into an amount we can work with to find out how much acid is required for a complete reaction. The steps taken to calculate the moles of NaOH lay the groundwork for determining the volumes of different acid solutions needed for neutralization in an acid-base titration.

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Most popular questions from this chapter

What volume of each of the following bases will react completely with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) ? a. \(0.100 \mathrm{M} \mathrm{NaOH}\) b. \(0.0500 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) c. \(0.250 \mathrm{M} \mathrm{KOH}\)

Assign the oxidation state for nitrogen in each of the following. a. \(\mathrm{Li}_{3} \mathrm{~N}\) b. \(\mathrm{NH}_{3}\) c. \(\mathrm{N}_{2} \mathrm{H}_{4}\) d. \(\mathrm{NO}\) e. \(\mathrm{N}_{2} \mathrm{O}\) f. \(\mathrm{NO}_{2}\) g. \(\mathrm{NO}_{2}^{-}\) h. \(\mathrm{NO}_{3}^{-}\) I. \(\mathrm{N}_{2}\)

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of barium chloride is mixed with \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of iron(III) sulfate?

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 121 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

A \(100.0\) -mLaliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) 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.
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