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Chapter 4: Problem 103

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a \(42.3 \mathrm{~mL}\) aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

Short Answer

Expert verified
The molarity of the caesium hydroxide solution is 0.852 M. The balanced chemical equation for the reaction between caesium hydroxide and hydroiodic acid is \(CsOH + HI \rightarrow CsI + H_2O\). The concentration of the hydroiodic acid solution is 0.376 M.

Step by step solution

01

Calculate the molarity of the CsOH solution

First, we need to determine the molarity of the CsOH solution. To do this, we'll follow these steps: 1. Determine the number of moles of CsOH in the solution 2. Calculate the molarity using the volume of the solution Number of moles of CsOH can be calculated by dividing the mass of CsOH by its molar mass: Moles of CsOH = mass / molar mass The mass of CsOH provided is 3.20 g. The molar mass of CsOH (Cs = 132.91 g/mol, O = 16.00 g/mol, H = 1.01 g/mol) is 149.92 g/mol. So the number of moles of CsOH is: moles = (3.20 g) / (149.92 g/mol) = 0.0213 mol Now, use the volume of the solution to calculate the molarity: Molarity = moles / volume The volume of the solution is 25.00 mL, which should be converted to liters: 25.00 mL * (1 L / 1000 mL) = 0.025 L The molarity of the CsOH solution is: Molarity = (0.0213 mol) / (0.025 L) = 0.852 M
02

Write the balanced chemical equation

Now we need to write the balanced chemical equation for the reaction between the caesium hydroxide (CsOH) and hydroiodic acid (HI) solutions. The reaction is a typical acid-base reaction, where the hydrogen ion (H鈦�) from the acid (HI) combines with the hydroxide ion (OH鈦�) from the base (CsOH) to form water (H鈧侽), and the remaining ions form a salt, in this case, caesium iodide (CsI). Here is the balanced chemical equation: CsOH + HI 鈫� CsI + H鈧侽
03

Calculate the molarity of the HI solution

Given the volume of the CsOH solution required to neutralize an aliquot of the HI solution, we can now calculate the molarity of the HI solution. Since the balanced chemical equation (Step 2) has a 1:1 stoichiometry between CsOH and HI, we can use the volume and molarity of the CsOH solution to determine the moles of HI that reacted and, subsequently, the concentration (molarity) of the HI solution. First, calculate the moles of CsOH in the titration: moles CsOH = molarity * volume We're given that 18.65 mL of the 0.852 M CsOH solution were used, so we convert this volume to L (18.65 mL * (1 L / 1000 mL) = 0.01865 L) and calculate the moles of CsOH: moles CsOH = (0.852 mol/L) * (0.01865 L) = 0.0159 mol Since the stoichiometry between CsOH and HI is 1:1, the moles of HI must be the same as the moles of CsOH: 0.0159 mol Now, use the volume of the HI aliquot to calculate its molarity: molarity HI = moles HI / volume HI The volume of the HI aliquot is given as 42.3 mL, which we convert to L (42.3 mL * (1 L / 1000 mL) = 0.0423 L). Then, calculate the HI molarity: molarity HI = (0.0159 mol) / (0.0423 L) = 0.376 M So the concentration of the hydroiodic acid solution is 0.376 M.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs in a reaction when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of the reactants and products over time. It is important to understand that equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their concentrations remain constant.

In the context of acid-base reactions, like the one between caesium hydroxide (CsOH) and hydroiodic acid (HI), reaching equilibrium usually involves complete neutralization unless specified otherwise. This means that every molecule of the base reacts with a molecule of the acid until neither reactant is in excess.

For students learning about equilibrium, it's crucial to recognize this balance and think about how changing concentrations of either reactant could shift equilibrium, as described by Le Chatelier's Principle. However, in titration exercises like this, equilibrium is typically reached by complete neutralization at the endpoint.
Balanced Chemical Equation
A balanced chemical equation is crucial in any chemical reaction because it provides a precise depiction of what happens during the reaction. It tells us the exact number of molecules or moles of each reactant and product, ensuring the conservation of mass based on the Law of Conservation of Mass.

In our CsOH and HI reaction, the balanced chemical equation is:
  • CsOH + HI 鈫� CsI + H鈧侽
This equation shows that one mole of cesium hydroxide reacts with one mole of hydroiodic acid to yield one mole of cesium iodide and one mole of water. This 1:1 stoichiometry is essential when using titration methods to determine unknown concentrations.

Chemical equations must be balanced for both atoms and charge, but in this case, the reaction is straightforward because we are dealing with ionic species in aqueous solution, where the ions combine simply to neutralize each other and form stable substances.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It provides a method to calculate the quantities of reactants or products based on the balanced chemical equation.

In the exercise provided, the primary focus is on calculating the molarity, which is a measure of concentration in terms of moles of solute per liter of solution. Molarity is necessary for determining the stoichiometry of the reaction, particularly in titrations.

Given the 1:1 stoichiometry between CsOH and HI, we used the molarity of CsOH and its volume to find the moles of HI required to reach the neutralization point. By converting the volumes from milliliters to liters and using molarity formulas, stoichiometry helps us to calculate the unknown concentration of the HI solution.

Understanding stoichiometry empowers you to predict the outcomes of reactions and to prepare solutions with desired concentrations, which is a fundamental skill in chemistry.

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Most popular questions from this chapter

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), should you dissolve in water to make \(1.00 \mathrm{~L}\) of vodka (which is an aqueous solution that is \(6.86 \mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{~g} / \mathrm{mL})\), calculate the volume of ethanol you need to make \(1.00 \mathrm{~L}\) of vodka.

A person suffering from hyponatremia has a sodium ion concentration in the blood of \(0.118 \mathrm{M}\) and a total blood volume of \(4.6 \mathrm{~L}\). What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to \(0.138 \mathrm{M}\), assuming no change in blood volume?

(a) What volume of \(0.115 \mathrm{MHClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{MNaOH}\) ? (b) What volume of \(0.128 \mathrm{MHCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\) ), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of a 0.108 \(M\) HCl solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of KOH must be present in the solution?

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

You make \(1.000 \mathrm{~L}\) of an aqueous solution that contains \(35.0 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) (a) What is the molarity of sucrose in this solution? (b) How many liters of water would you have to add to this solution to reduce the molarity you calculated in part (a) by a factor of two?
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