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Hydrochloric acid (75.0 mL of 0.250 \(M\) ) is added to 225.0 mL of 0.0550 \(M\) Ba(OH) \(_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

Short Answer

Expert verified
The concentration of the excess OH鈦 ions left in the solution is \(0.020\, M\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)鈧) is: \[ 2\,\mathrm{HCl}\ +\ \mathrm{Ba(OH)_{2}}\ ->\ 2\,\mathrm{H_{2}O}\ +\ \mathrm{BaCl_{2}} \]
02

Calculate the moles of each reactant

For HCl: Moles = Volume 脳 Molarity Moles \(= (0.250\, M) \times (0.075\, L)\) Moles \(= 0.01875\, mol\) For Ba(OH)鈧: Moles = Volume 脳 Molarity Moles \(= (0.0550\, M) \times (0.225\, L)\) Moles \(= 0.012375\, mol\)
03

Determine the limiting reactant

Compare the ratio of moles to the stoichiometric coefficients in the balanced chemical equation: \(0.01875\, mol\, \mathrm{HCl}\) / 2 \(< 0.012375\, mol\, \mathrm{Ba(OH)_{2}} / 1\) Since the ratio for HCl is smaller, it is the limiting reactant.
04

Calculate moles of excess reactant left

From the balanced chemical equation, we know that two moles of HCl react with one mole of Ba(OH)鈧. Moles of Ba(OH)鈧 reacted = (0.01875 mol HCl) 脳 (1 mol Ba(OH)鈧 / 2 mol HCl) Moles of Ba(OH)鈧 reacted = 0.009375 mol Moles of Ba(OH)鈧 left = Initial moles Ba(OH)鈧 - Moles of Ba(OH)鈧 reacted Moles of Ba(OH)鈧 left = 0.012375 - 0.009375 = 0.003 mol
05

Calculate the concentration of the excess OH鈦 ions

Each mole of Ba(OH)鈧 contributes 2 moles of OH鈦 ions, so we have: Moles of OH鈦 left = 2 脳 Moles of Ba(OH)鈧 left = 2 脳 0.003 mol = 0.006 mol The total volume of the solution after the reaction is: Total volume = Volume of HCl + Volume of Ba(OH)鈧 = 0.075 L + 0.225 L = 0.300 L Finally, the concentration of excess OH鈦 ions in the solution: Concentration of OH鈦 = Moles of OH鈦 left / Total volume Concentration of OH鈦 \(= 0.006\, mol / 0.300\, L = 0.020\, M\) So, the concentration of the excess OH鈦 ions left in the solution is \(0.020\, M\).

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Key Concepts

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

Understanding Acid-Base Reactions
An acid-base reaction is a chemical process where an acid donates protons (H+) to a base. In the context of titration, this is a pivotal concept, as the goal is to determine the unknown concentration of an acid or base by reacting it with a solution of known concentration. Typically, the acid and base react in a predictable stoichiometry, such as the neutralization reaction where HCl reacts with Ba(OH)2 producing water and a salt, BaCl2. The reaction shows how the H+ ions from the acid combine with the OH- ions from the base, resulting in the production of water.

It's essential to start titration calculations by writing out the balanced chemical equation, as this sets the stage for all subsequent calculations and enables you to visualize the reactants and products involved. This understanding is the gateway to mastering titrations and is critical for processing the related calculations confidently.
Stoichiometry in Action
Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It's about balancing equations and using these balances to calculate different aspects of the chemical reaction. For instance, the balanced equation for the reaction between HCl and Ba(OH)2 involves a 2:1 stoichiometry, indicating that two moles of HCl are required to completely react with one mole of Ba(OH)2.

In an acid-base titration, stoichiometry allows us to predict the amounts of reactants that will be consumed and to determine the endpoint of the titration. By knowing the stoichiometric relationships, we can calculate precisely how much acid or base will be needed to neutralize the other, which is a fundamental aspect of titration calculations.
Identifying the Limiting Reactant
The limiting reactant in a chemical reaction is the substance that is completely consumed first and thus determines the amount of product that can be formed. This concept is a key aspect of titration calculations because it dictates when the reaction will stop. By determining the limiting reactant, you can calculate the moles of excess reactant as well as the final concentration of ions remaining in solution.

For any given titration, comparing the mole ratio of the reactants with the stoichiometric coefficients in the balanced chemical equation enables you to identify the limiting reactant. This step is essential as it establishes the foundation for further calculations, such as determining the moles of excess reactant and the concentration of remaining ions.
Calculating Molarity in Solutions
Molarity is a measure of the concentration of a solute in a solution and is expressed in moles per liter (M). It's a critical concept in titration calculations because it helps to quantify the amount of acid or base in a given volume of solution. By knowing the molarity, we can calculate the moles of the solute by multiplying the molarity by the volume of the solution in liters.

In the case of our exercise, once the reaction has been completed, we calculate the molarity of the excess ions remaining in solution. This involves dividing the moles of excess OH- ions by the total volume of the solution after the reaction. Understanding how to manipulate molarity, volume, and moles is fundamental for accurate titration calculations and helps ensure a successful analysis of the chemical reaction under study.

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Most popular questions from this chapter

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second (L/s) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm}\) HCl into the stream. (See Exercise 123 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of 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: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ What mass of \(\mathrm{CaO}\) is consumed in an 8.00 -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.

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

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