Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For solutions of a weak acid, a graph of \(\mathrm{pH}\) versus the logarithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

Short Answer

Expert verified
The magnitude of the slope of the graph of pH versus the logarithm of the initial acid concentration for solutions of weak acids is \( \frac{1}{2} \).

Step by step solution

01

Write the ionization equation for a weak acid

Let's denote the weak acid as "HA". The ionization of a weak acid in water can be represented as: \[HA + H_2O \rightleftharpoons H_3O^+ + A^-\]
02

Write the expression for the ionization constant

The ionization constant, Ka, represents the equilibrium concentrations of the ions, which can be calculated as: \[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \]
03

Simplify the expression for the concentrations

If we denote 'x' as the concentration of \(H_3O^+\), then \(A^-\) is also 'x' (since one HA produces one \(H_3O^+\) and one \(A^-\)). Thus, the concentration of HA remaining will be ([HA_initial] - x). Now we can rewrite the ionization constant expression as: \[K_a = \frac{x^2}{[HA_{initial}] - x}\]
04

Find the H3O+ concentration, x, in terms of Ka and the initial concentration of HA

Since the acid is weak, its ionization is small, and therefore we can assume that "x" is significantly smaller than the initial concentration of "HA": \[K_a = \frac{x^2}{[HA_{initial}]}\] Now we can solve for x: \[x = [H_3O^+] = \sqrt{K_a[HA_{initial}]}\]
05

Write the expression for pH and take the logarithm of both sides

The pH is related to the concentrations of the ions by the equation: \[ pH = -\log{[H_3O^+]}\] Substituting the expression we derived for x: \[pH = -\log{(\sqrt{K_a[HA_{initial}]})}\] Taking the logarithm of both sides, we get: \[\log{(pH)} = \log{(-(1/2))} \cdot (\log{(K_a)}) + (\log{([HA_{initial}] - \log{K_a})})\]
06

Find the magnitude of the slope

In the last expression, we can observe that the coefficient, '-(1/2)', is the slope of the graph. So, the magnitude of the slope is: \[ Slope = |- (\frac{1}{2})| = \frac{1}{2}\] Thus, the magnitude of the slope of the graph of pH versus the logarithm of the initial acid concentration for solutions of weak acids is 1/2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ionization constant (Ka)
The ionization constant, commonly known as the acid dissociation constant (K_a), is a vital concept when studying weak acids. It measures how much an acid separates into ions in a solution. This is usually represented in equilibrium conditions. For a weak acid, not all of it will dissociate into ions, which is different from strong acids that fully ionize.
For a given weak acid (HA), the ionization can be represented by the equation:
\[HA + H_2O \rightleftharpoons H_3O^+ + A^-\]
Using this equation, we write the ionization constant expression as:
\[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \]
Here, [K_a] gives insight into the strength of an acid. A higher K_a value indicates a stronger acid, which ionizes more in water. Conversely, a lower K_a signals a weaker acid as less ions are formed. Understanding how to use K_a is crucial for predicting the behavior of weak acid solutions.
pH calculation
Calculating the pH of a weak acid is a key skill for understanding its acidity. The pH is a logarithmic scale used to specify the acidity or basicity of a solution. It is determined by the concentration of hydronium ions (H_3O^+) in the solution.
For weak acids, because they do not ionize completely, calculating pH involves a bit more work. From the ionization constant expression, we can derive the concentration of H_3O^+:
\[x = [H_3O^+] = \sqrt{K_a[HA_{initial}]}\]
With [H_3O^+] known, the pH of the solution is calculated as:
\[ pH = -\log{[H_3O^+]}\]
This equation reflects the negative logarithm of the H_3O^+ concentration. A lower pH value indicates higher acidity. For weak acids, pH values are typically higher than for strong acids due to incomplete ionization.
Equilibrium concentrations
Understanding equilibrium concentrations is crucial for analyzing weak acid solutions. When a weak acid is in a solution, it establishes an equilibrium between the ionized and non-ionized forms.
In the ionization reaction:
\[HA + H_2O \rightleftharpoons H_3O^+ + A^-\]
The concentrations of products (H_3O^+ and A^-) and reactants (HA) fluctuate until they reach a state where they do not change with time. These are the equilibrium concentrations.
Using the initial concentration of the weak acid ([HA_{initial}]), we can determine the changes in concentration due to ionization using x:
  • Decrease in HA concentration by x: [HA] = [HA_{initial}] - x.
  • Increase in H_3O^+ by x: [H_3O^+] = x.
  • Increase in A^- by x: [A^-] = x.
A simplifying assumption often used is that x is much smaller than the initial concentration [HA_{initial}], especially for weak acids. This assumption allows us to easily estimate pH and the concentrations at equilibrium, making calculations more manageable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The structural formula for acetic acid is shown in Table 16.2 . Replacing hydrogen atoms on the carbon with chlorine atoms causes an increase in acidity, as follows: $$ \begin{array}{lll} \hline \text { Acid } & \text { Formula } & \boldsymbol{K}_{\boldsymbol{a}}\left(\mathbf{2 5}^{\circ} \mathbf{C}\right) \\\ \hline \text { Acetic } & \mathrm{CH}_{3} \mathrm{COOH} & 1.8 \times 10^{-5} \\\ \text {Chloroacetic } & \mathrm{CH}_{2} \mathrm{ClCOOH} & 1.4 \times 10^{-3} \\\ \text {Dichloroacetic } & \mathrm{CHCl}_{2} \mathrm{COOH} & 3.3 \times 10^{-2} \\\ \text {Trichloroacetic } & \mathrm{CCl}_{3} \mathrm{COOH} & 2 \times 10^{-1} \\\ \hline \end{array} $$ Using Lewis structures as the basis of your discussion, explain the observed trend in acidities in the series. Calculate the \(\mathrm{pH}\) of a \(0.010 \mathrm{M}\) solution of each acid.

(a) Give the conjugate base of the following Bronsted- Lowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}^{+}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-}\) (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{H}^{+}\right]=0.0505 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} \mathrm{M} ;(\mathrm{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},\) (b) \(\mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-},(\mathbf{c})\) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\)

Using data from Appendix \(\mathrm{D},\) calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following solutions: (a) \(0.10 \mathrm{M} \mathrm{NaBrO},\) (b) \(0.080 \mathrm{MNaHS}\), (c) a mixture that is \(0.10 \mathrm{M}\) in \(\mathrm{NaNO}_{2}\) and \(0.20 \mathrm{M}\) in \(\mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free