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Chapter 16: Problem 53

A 0.100\(M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is 11.0\(\%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right],\) and \(K_{a}\) for chloroacetic acid.

Short Answer

Expert verified
The equilibrium concentrations for ClCH鈧侰OO鈦�, H鈦�, and ClCH鈧侰OOH are found as follows: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right]_{eq}=0.0110 M\] \[\left[\mathrm{H}^{+}\right]_{eq}=0.0110 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right]_{eq}=0.0890 M\] The Ka value for chloroacetic acid is calculated as: \[K_{a}=1.36 \times 10^{-3}\]

Step by step solution

01

Calculate the ionization concentrations

From the given information, we know that 11% of chloroacetic acid ionizes. We can use this information to calculate the equilibrium concentrations of ClCH鈧侰OO鈦� and H鈦�. If the 0.1M solution is 11.0% ionized, then: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right]_{eq} = \left[\mathrm{H}^{+}\right]_{eq} = 0.11 \times 0.100 M = 0.0110 M\]
02

Calculate the equilibrium concentration of chloroacetic acid

Next, let's determine the equilibrium concentration of ClCH鈧侰OOH. Since 11% of it ionized, then 100% - 11% = 89% of it remains un-ionized in the equilibrium: \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right]_{eq} = 0.89 \times 0.100 M = 0.0890 M\]
03

Calculate the acid dissociation constant (Ka)

Now that we have the equilibrium concentrations, we can use them to find the Ka of chloroacetic acid. The ionization dissociation reaction of chloroacetic acid can be written as: \[\mathrm{ClCH}_{2}\mathrm{COOH} \rightleftharpoons \mathrm{ClCH}_{2}\mathrm{COO}^{-}+\mathrm{H}^{+}\] The Ka expression for this reaction is: \[K_{a}=\frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]} \] Inserting the calculated equilibrium concentrations: \[K_{a}=\frac{(0.0110)(0.0110)}{(0.0890)} = 1.36 \times 10^{-3}\] We have obtained the equilibrium concentrations for ClCH鈧侰OO鈦�, H鈦� and ClCH鈧侰OOH, as well as the Ka value. The results are as follows: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right]_{eq}=0.0110 M\] \[\left[\mathrm{H}^{+}\right]_{eq}=0.0110 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right]_{eq}=0.0890 M\] \[K_{a}=1.36 \times 10^{-3}\]

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Key Concepts

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

Ionization of Acids
When we talk about acids ionizing, we are referring to the process where acid molecules disassociate into ions in water. This process is crucial because it directly affects how the acid behaves in a solution. For weak acids like chloroacetic acid, only a fraction will ionize.

In the provided exercise, chloroacetic acid ionizes to produce chloride acetate ions \([\mathrm{ClCH}_2\mathrm{COO}^-]\) and hydrogen ions \([\mathrm{H}^+]\). This percentage of ionization is significant because it gives us a measure of the acid's strength: the higher the percentage of ionization, the stronger the acid.

Key points about ionization of acids include:
  • Ionization can vary: different acids have different ionization levels, depending on their chemical structure.
  • Concentration matters: higher initial concentrations can lead to different equilibrium states.
  • Equilibrium form: only a part of a weak acid ionizes, with the rest remaining in its original form, affecting both concentration and pH level.
Understanding these concepts helps you predict the behavior of acids during reactions.
Equilibrium Concentrations
Equilibrium concentrations are the amounts of each species present in a reaction mixture when the reaction has reached equilibrium. At this point, the rates of the forward and backward reactions are equal, so the concentrations stay constant.

In the exercise, we calculate these equilibrium concentrations based on the ionization percentage of chloroacetic acid. Here's what you should know:
  • Equilibrium and starting concentrations: The initial concentration tells us how much acid was present before any ionization occurred. Here, we started with 0.100 M chloroacetic acid.
  • Ionization percentage impact: Knowing that 11% ionized, we determined the concentrations of the ions, which were both 0.0110 M.
  • Remaining concentration: The part of the acid that did not ionize remained in the solution, contributing to an equilibrium concentration of 0.0890 M for chloroacetic acid.
These steps are fundamental in understanding how much of each substance remains in a solution at equilibrium.
Chemical Equilibrium Calculations
Chemical equilibrium calculations allow us to quantify the concentrations of various species in a reaction mixture. This process often involves using an equilibrium constant, in this case, the acid dissociation constant (Ka).

The calculation for chloroacetic acid involved determining Ka, which is a quantitative measure of the strength of an acid. Let's break it down:
  • The formula: The general formula for Ka is \(K_{a}=\frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]} \), used here to relate the equilibrium concentrations.
  • Substituting values: We plug in the values calculated for equilibrium concentrations into the formula to find Ka.
  • Result interpretation: A calculated Ka value of \(1.36 \times 10^{-3}\) indicates the acid鈥檚 dissociation degree, helping predict behavior in different scenarios.
By mastering these calculations, students can better understand reaction dynamics and predict how changes in conditions could affect equilibrium.

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Most popular questions from this chapter

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

Identify the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each equation, and also identify the conjugate acid and conjugate base of each on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: \((\mathbf{a}) 0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{g}\) of \(\mathrm{KOH}\) in 500.0 mL of solution, ( c ) 10.0 \(\mathrm{mL}\) of 0.0105 \(\mathrm{MCa}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{mL},(\mathbf{d})\) a solution formed by mixing 20.0 \(\mathrm{mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 \(\mathrm{mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}.\)

Is each of the following statements true or false? (a) All strong acids contain one or more H atoms. (b) A strong acid is a strong electrolyte. (c) A \(1.0-M\) solution of a strong acid will have \(\mathrm{pH}=1.0 .\)

(a) Write a chemical equation that illustrates the auto-ionization of water. (b) Write the expression for the ion-product constant for water \(K_{w}\) . (c) If a solution is described as basic, which of the following is true: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right],\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)
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