/*! 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 96 The solubility of \(\mathrm{Cl}_... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 96

The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$ \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q) $$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\), calculate the equilibrium concentration of \(\mathrm{HClO}\) formed. (b) What is the \(\mathrm{pH}\) of the final solution?

Short Answer

Expert verified
The equilibrium concentration of HClO is 0.0077 M and the pH of the final solution is approximately 2.11.

Step by step solution

01

Calculate the initial concentrations of reactants and products

First, let's calculate the initial concentration of Cl鈧� in water. We know the solubility of Cl鈧� is 310 cm鲁 in 100 g of water. Since the molar volume at STP is 22.4 L/mol, we can calculate the initial moles of Cl鈧� and then use the volume of water to find the concentration. Moles of Cl鈧� = (310 cm鲁) / (22.4 L/mol x 1000 cm鲁/L) = 0.0138 mol Since the volume of water is 100 g and its density is 1 g/mL, we can find the volume in L: Volume of water = (100 g) / (1 g/mL) = 100 mL = 0.1 L The initial concentration of Cl鈧� is: [Cl鈧俔 = (0.0138 mol) / (0.1 L) = 0.138 M Initially, [H鈧侽] 鈮� 55.5 M, [Cl鈦籡 = [HClO] = [H鈦篯 = 0.
02

Set up the equilibrium expression and solve for [HClO]

We can now set up the equilibrium expression for the reaction. The given equilibrium constant K = 4.7 脳 10鈦烩伌. The reaction can be represented as: Cl鈧�(aq) + H鈧侽 鈬� Cl鈦�(aq) + HClO(aq) + H鈦�(aq) We can use an ICE table (Initial, Change, Equilibrium) to find the equilibrium concentrations of each species: Initial | Change | Equilibrium ------------------------------- Cl鈧� | 0.138 M | -x M | 0.138 - x H鈧侽 | 55.5 M | -x M | 55.5 - x Cl鈦� | 0 M | +x M | x HClO | 0 M | +x M | x H鈦� | 0 M | +x M | x The equilibrium expression is: K = ([Cl鈦籡[HClO][H鈦篯) / ([Cl鈧俔[H鈧侽]) We can plug the equilibrium concentrations and the value of K into this expression: 4.7 脳 10鈦烩伌 = (x * x * x) / ((0.138 - x)(55.5 - x)) As the value of K is very small, we can assume that x << 55.5 and x << 0.138. Thus, the equation becomes: 4.7 脳 10鈦烩伌 鈮� (x鲁) / (0.138 x 55.5) Now, we can solve for x, which represents the concentration of HClO at equilibrium.
03

Calculate the equilibrium concentration of HClO

Rearrange the previous equation to solve for x: x鲁 鈮� 4.7 脳 10鈦烩伌 脳 0.138 脳 55.5 x 鈮� (4.7 脳 10鈦烩伌 脳 0.138 脳 55.5)^(1/3) 鈮� 0.0077 M Thus, the equilibrium concentration of HClO is 0.0077 M.
04

Determine the pH of the final solution

Since the concentration of H鈦� ions at equilibrium is equal to x, we have: [H鈦篯 = x = 0.0077 M Now, we can determine the pH of the final solution: pH = -log[H鈦篯 = -log(0.0077) 鈮� 2.11 The pH of the final solution is approximately 2.11.

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Key Concepts

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

Solubility
Solubility is how much of a substance can dissolve in a solvent. For chlorine gas (\(\text{Cl}_2\)), its solubility in water can be measured under standard conditions. At standard temperature and pressure (STP), the solubility of \(\text{Cl}_2\) in water is given as 310 cm鲁 per 100 g of water.
This means that under these conditions, 310 cm鲁 of chlorine gas can dissolve in 100 g of water, which is very useful information when calculating concentrations in chemical reactions.
To find out how much of \(\text{Cl}_2\) is dissolved in terms of moles, you need to know the volume of the gas and the volume it occupies at STP, which is 22.4 liters per mole.
Using these values, we can calculate the number of moles of \(\text{Cl}_2\) in the solution, which helps in further calculations regarding chemical equilibrium and pH.
Chemical Equilibrium
Chemical equilibrium is a key concept in chemical reactions. It occurs when a reaction and its reverse happen at the same rate. Let's consider the equilibrium reaction with chlorine in water: \(\text{Cl}_2(aq) + \text{H}_2\text{O} ightleftharpoons \text{Cl}^-(aq) + \text{HClO}(aq) + \text{H}^+(aq)\). This means the forward and reverse reactions happen at the same speed, and the concentrations of reactants and products remain constant.
You can calculate equilibrium concentrations using an ICE (Initial, Change, Equilibrium) table. The equilibrium constant, \(K\), reflects how far a reaction goes to completion at equilibrium. For this reaction, \(K = 4.7 \times 10^{-4}\), indicating the reaction favors the reactants under these conditions.
By setting up the equilibrium expression: \(K = \frac{[\text{Cl}^-][\text{HClO}][\text{H}^+]}{[\text{Cl}_2][\text{H}_2\text{O}]}\), and solving for the concentrations of the products at equilibrium, we better understand the reaction's dynamics.
pH Calculation
The pH of a solution helps us understand its acidity or alkalinity. pH is defined as the negative logarithm of the hydrogen ion concentration, \(-\log[\text{H}^+]\). In the chlorine-water equilibrium reaction, the concentration of \(\text{H}^+\) ions is equal to \(x\) from the ICE table, calculated to be \(0.0077\,\text{M}\).
This is because for each mole of \(\text{Cl}_2\) that reacts, one mole of \(\text{H}^+\) is produced.Calculating the pH involves:
  • Taking the negative log of the \(\text{H}^+\) concentration.
  • In our case, \(\text{pH} = -\log(0.0077) \approx 2.11\).
This result indicates the solution is acidic, as a pH less than 7 means more hydrogen ions compared to water.

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Most popular questions from this chapter

Account for the following observations: (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) are quite basic.

Explain why \(\mathrm{SO}_{2}\) can be used as a reducing agent but \(\mathrm{SO}_{3}\) cannot.

Write a balanced equation for the preparation of \(\mathrm{H}_{2}\) using (a) \(\mathrm{Mg}\) and an acid, (b) carbon and steam, (c) methane and steam.

The half-life of cobalt-60 is \(5.27\) yr. How much of a \(1.000-\mathrm{mg}\) sample of cobalt-60 is left after \(15.81 \mathrm{yr}\) ? SOLUTION Analyze We are given the half-life for cobalt-60 and asked to calculate the amount of cobalt-60 remaining from an initial \(1.000\)-mg sample after \(15.81\) yr. Plan We will use the fact that the amount of a radioactive substance decreases by \(50 \%\) for every half-life that passes. Solve Because \(5.27 \times 3=15.81,15.81\) yr is three half-lives for cobalt-60. At the end of one half-life, \(0.500 \mathrm{mg}\) of cobalt-60 remains, \(0.250 \mathrm{mg}\) at the end of two half-lives, and \(0.125 \mathrm{mg}\) at the end of three half-lives.

(a) The \(\mathrm{P}_{4}, \mathrm{P}_{4} \mathrm{O}_{6}\) and \(\mathrm{P}_{4} \mathrm{O}_{10}\) molecules have a common structural feature of four \(\mathrm{P}\) atoms arranged in a tetrahedron (Figures \(22.27\) and 22.28). Does this mean that the bonding between the \(\mathrm{P}\) atoms is the same in all these cases? Explain. (b) Sodium trimetaphosphate \(\left(\mathrm{Na}_{3} \mathrm{P}_{3} \mathrm{O}_{9}\right)\) and sodium tetrametaphosphate \(\left(\mathrm{Na}_{4} \mathrm{P}_{4} \mathrm{O}_{12}\right)\) are used as water-softening agents. They contain cyclic \(\mathrm{P}_{3} \mathrm{O}_{9}{ }^{3-}\) and \(\mathrm{P}_{4} \mathrm{O}_{12}{ }^{4-}\) ions, respectively. Propose reasonable structures for these ions.
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