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Chapter 17: Problem 83

A monoprotic acid HX has \(K_{\mathrm{a}}=1.3 \times 10^{-3} .\) Calculate the equilibrium concentrations of \(\mathrm{HX}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{pH}\) for a \(0.010 \mathrm{M}\) solution of the acid.

Short Answer

Expert verified
Equilibrium [HX] = 0.0064 M, [H鈦篯 = 0.0036 M, pH = 2.44.

Step by step solution

01

Write the Equilibrium Expression

For the dissociation of a monoprotic acid \( HX \) in water, the chemical equation is \( HX \rightleftharpoons H^+ + X^- \). The expression for the equilibrium constant \( K_a \) is given by:\[ K_a = \frac{[H^+][X^-]}{[HX]} \] Where \( K_a = 1.3 \times 10^{-3} \) and the initial concentration of \( [HX] = 0.010 M \).
02

Identify the Initial and Equilibrium Concentrations

Initially, the concentration of \( [HX] \) is \( 0.010 M \), and concentrations of \( [H^+] \) and \( [X^-] \) are 0. At equilibrium, let \( x \) be the change in concentration of \( [H^+] \) and \( [X^-] \), so:- \( [HX] = 0.010 - x \)- \( [H^+] = x \)- \( [X^-] = x \)
03

Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_a = \frac{x \cdot x}{0.010 - x} = \frac{x^2}{0.010 - x} \] Given \( K_a = 1.3 \times 10^{-3} \), the equation becomes:\[ 1.3 \times 10^{-3} = \frac{x^2}{0.010 - x} \]
04

Approximate and Solve the Quadratic Equation

Assume that \( x \) is small compared to 0.010, so \( 0.010 - x \approx 0.010 \). Thus, the equation simplifies to:\[1.3 \times 10^{-3} = \frac{x^2}{0.010} \] Canceling 0.010 from both sides gives:\[x^2 = 1.3 \times 10^{-5} \]Solve for \( x \):\[x = \sqrt{1.3 \times 10^{-5}} \approx 0.0036 \]
05

Calculate Equilibrium Concentrations and pH

Given \( x = 0.0036 \), the equilibrium concentrations are:- \( [H^+] = 0.0036 \) M- \( [HX] = 0.010 - 0.0036 = 0.0064 \) MTo find the pH:\[ \text{pH} = -\log[H^+] = -\log(0.0036) \approx 2.44\]

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Key Concepts

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

Equilibrium Constant (Ka)
The equilibrium constant, known as the acid dissociation constant \(K_a\), is a fundamental concept in chemistry. It tells us how strong a monoprotic acid is by measuring how well it dissociates in solution. For a given acid \(HX\), the equilibrium expression is
\[K_a = \frac{[H^+][X^-]}{[HX]}\]The value of \(K_a\) is specific to each acid and indicates how much the acid will ionize in water. A large \(K_a\) means the acid dissociates well and is strong, while a small \(K_a\) indicates a weaker acid.
  • If \(K_a\) is high, the equilibrium lies to the right, favoring products \([H^+][X^-]\).
  • If \(K_a\) is low, the equilibrium lies to the left, favoring the undissociated form \([HX]\).
Dissociation of Acids
Dissociation is the process by which an acid breaks apart into ions when dissolved in water. For monoprotic acids like \(HX\), this involves the release of one proton \((H^+)\). The chemical equation representing the dissociation can be written as:
\[HX \rightleftharpoons H^+ + X^-\]Initially, we start with a concentration of the acid, while the concentrations of \(H^+\) and \(X^-\) are zero. As the acid dissociates, some of \(HX\) becomes \(H^+\) and \(X^-\), reaching an equilibrium state.
  • The degree of dissociation is related to \(K_a\). Higher \(K_a\) means more dissociation.
  • Understanding initial and equilibrium concentrations helps in calculating how much of each species is present at equilibrium.
pH Calculation
The pH is a scale used to specify the acidity or basicity of an aqueous solution, and is calculated using the formula:
\[\text{pH} = -\log[H^+]\]To find the pH of a solution, you need the concentration of \(H^+\) ions, which you can determine from the equilibrium concentrations.
In our example, after calculating \(x\), the equilibrium concentration of \(H^+\) is found to be 0.0036 M. Using this value:
\[\text{pH} = -\log(0.0036) \approx 2.44\]This tells us that our solution is acidic, as pH values less than 7 are considered acidic.

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Most popular questions from this chapter

Which of the following common food additives would give a basic solution when dissolved in water? (a) \(\mathrm{NaNO}_{3}\) (used as a meat preservative) (b) \(\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\) (sodium benzoate; used as a soft-drink preservative) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) (used as an emulsifier in the manufacture of pasteurized cheese)

The nickel(II) ion exists as \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) in aqueous solution. Why is this solution acidic? As part of your answer, include a balanced equation depicting what happens when \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) interacts with water.

Arrange the following \(0.10 \mathrm{M}\) solutions in order of increasing pH. (a) \(\mathrm{NaCl}\) (b) \(\mathrm{NH}_{4} \mathrm{Cl}\) (c) \(\mathrm{HCl}\) (d) \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\) (e) \(\mathrm{KOH}\)

Chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}\), is a moderately weak acid \(\left(K_{\mathrm{a}}=1.40 \times 10^{-3}\right) .\) If you dissolve \(94.5 \mathrm{mg}\) of the acid in water to give 125 mL of solution, what is the \(\mathrm{pH}\) of the solution?

Given the following solutions: (a) \(0.1 \mathrm{M} \mathrm{NH}_{3}\) (b) \(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.1 \mathrm{M} \mathrm{NaCl}\) (d) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (e) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (f) \(0.1 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{CO}_{2}\) (g) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) (i) Which of the solutions are acidic? (ii) Which of the solutions are basic? (iii) Which of the solutions is most acidic?
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