/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 137 A 0.500 -L sample of \(\mathrm{H... [FREE SOLUTION] | 91影视

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A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{mL}\) of \(0.213 \mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{mL}\) of \(0.103 M\) HCI for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

Short Answer

Expert verified
The molarity of the original H鈧係O鈧 solution is 0.00929 M.

Step by step solution

01

Find moles of OH鈦 ions neutralized by excess HCl

We know that: Volume of HCl = 13.21 mL Molarity of HCl = 0.103 M Let's convert the volume to liters and then use the molarity formula to calculate moles of OH鈦 ions in the excess base. \(Volume_{HCl} (L) = \frac{13.21~mL}{1000} = 0.01321 L\) Moles of OH鈦 ions = Molarity 脳 Volume Moles of OH鈦 ions = 0.103 M 脳 0.01321 L Moles of OH鈦 ions = 0.00136 mol
02

Find total moles of OH鈦 ions added to H鈧係O鈧

We know that: Volume of NaOH = 50.0 mL Molarity of NaOH = 0.213 M Convert the volume to liters and then use the molarity formula to calculate moles of OH鈦 ions added to H鈧係O鈧. \(Volume_{NaOH} (L) = \frac{50.0~mL}{1000} = 0.05 L\) Moles of OH鈦 ions = Molarity 脳 Volume Moles of OH鈦 ions = 0.213 M 脳 0.05 L Moles of OH鈦 ions = 0.01065 mol
03

Calculate moles of H鈦 ions in H鈧係O鈧

First, let's find the moles of OH鈦 ions that reacted with H鈧係O鈧 by subtracting the moles of OH鈦 ions that reacted with the excess acid from the total moles of OH鈦 ions added. Moles of OH鈦 ions reacted with H鈧係O鈧 = Total moles of OH鈦 ions - Moles of OH鈦 ions in excess base Moles of OH鈦 ions reacted with H鈧係O鈧 = 0.01065 mol - 0.00136 mol Moles of OH鈦 ions reacted with H鈧係O鈧 = 0.00929 mol Since H鈧係O鈧 has two acidic hydrogens, it reacts with two moles of OH鈦 ions for every mole of H鈧係O鈧. Therefore, the moles of H鈧係O鈧 can be calculated as: Moles of H鈧係O鈧 = Moles of OH鈦 ions reacted with H鈧係O鈧 梅 2 Moles of H鈧係O鈧 = 0.00929 mol 梅 2 Moles of H鈧係O鈧 = 0.004645 mol
04

Calculate the molarity of the original H鈧係O鈧 solution

We are given that the volume of the original H鈧係O鈧 solution is 500 mL. Let's calculate the molarity of the solution using the formula: Molarity = moles of solute 梅 volume of solution (in L) First, convert the volume to liters: \(Volume_{H_2SO_4} (L) = \frac{500~mL}{1000} = 0.5 L\) Now, calculate the molarity of the solution: Molarity of H鈧係O鈧 = 0.004645 mol 梅 0.5 L Molarity of H鈧係O鈧 = 0.00929 M The molarity of the original H鈧係O鈧 solution is 0.00929 M.

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a fundamental process in chemistry used to determine the concentration of an unknown acid or base by neutralizing it with a base or acid of known concentration, respectively. The endpoint of the titration is reached when the number of moles of hydrogen ions, H鈦, in the acid equals the number of moles of hydroxide ions, OH鈦, from the base, signaling complete neutralization.

Here's the basic premise: A carefully measured volume of a standard solution (the titrant with known molarity) is added to a known volume of the solution with the unknown concentration until the reaction reaches the endpoint. This is indicated by a color change from an indicator or by reaching a specified pH value using a pH meter. In our exercise, we can infer the molarity of the sulfuric acid solution by neutralizing it with a known quantity of sodium hydroxide (NaOH) and then back-titrating the excess NaOH with hydrochloric acid (HCl).

Molarity
Molarity, symbolized as M, is a unit of concentration within chemistry that measures the number of moles of a solute per liter of solution. To calculate molarity, you use the formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). For our problem, this concept allows us to find the molarity of sulfuric acid by determining the moles of H鈧係O鈧 and dividing it by the total volume of the solution in which H鈧係O鈧 has dissolved. This calculation helps to understand the strength of the acid solution and plays a crucial role in various applications, including chemical reactions and titration exercises like the one we explored.

Stoichiometry
Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in a given reaction. In acid-base reactions, stoichiometry helps us to understand the mole-to-mole ratios between acid and base. For sulfuric acid, which has a stoichiometric factor of two due to its two acidic hydrogens, it will react with two moles of hydroxide ions for each mole of H鈧係O鈧.

By using stoichiometry, we can convert the volume and concentration (molarity) of NaOH added to moles, determine the moles of OH鈦 involved in the reaction, and hence calculate the moles of H鈧係O鈧 in the sample. Understanding stoichiometry is pivotal for accurate chemical calculations and is a cornerstone of high-quality chemistry education.

Sulfuric Acid (H鈧係O鈧)
Sulfuric acid is a strong acid widely used in industry and laboratories. It's characterized by its chemical formula H鈧係O鈧 and is known for having two acidic protons, which means it can donate two H鈦 ions per molecule in reactions. In the context of our titration exercise, recognizing that sulfuric acid can react with two moles of base per mole of acid is essential to accurately perform calculations.

Given its nature, when it dissociates in water, it contributes to the acid's molarity based on the relation that 1 M of sulfuric acid provides 2 M of H鈦 ions, demonstrating why stoichiometry is a critical factor in getting the correct molarity. When dealing with sulfuric acid in calculations and experimental setups, always consider its dual acidic nature for precise results.

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Most popular questions from this chapter

A 450.0 -mL sample of a 0.257 - \(M\) solution of silver nitrate is mixed with \(400.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{mL}\) of a \(0.100-M\) solution of \(\mathrm{AgNO}_{3} ?\)

Some of the substances commonly used in stomach antacids are \(\mathrm{MgO}, \mathrm{Mg}(\mathrm{OH})_{2},\) and \(\mathrm{Al}(\mathrm{OH})_{3}\) a. Write a balanced equation for the neutralization of hydrochloric acid by each of these substances. b. Which of these substances will neutralize the greatest amount of 0.10 \(M\) HCl per gram?

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\operatorname{FeSO}_{4}(a q)+\operatorname{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of \(204.22 \mathrm{g} / \mathrm{mol}\). In the titration, \(34.67 \mathrm{mL}\) of the sodium hydroxide solution was required to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.

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