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

The pKa of a weak acid is \(4.8 .\) What should be the ratio of [acid] \(/[\mathrm{salt}]\), if a buffer of \(\mathrm{pH}=5.8\) is required? (a) \(0.1\) (b) 10 (c) 1 (d) 2

Short Answer

Expert verified
The ratio of [acid]/[salt] should be 0.1, so the answer is (a) 0.1.

Step by step solution

01

Understand the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer solution, consisting of a weak acid and its salt (conjugate base), is given by: \[ \text{pH} = \text{pKa} + \log\left( \frac{[\text{salt}]}{[\text{acid}]} \right) \]This equation helps us calculate the pH of a buffer solution based on the concentration ratio of salt to acid.
02

Substitute Given Values

We are given that the pH of the buffer is 5.8 and the pKa of the weak acid is 4.8. Substitute these values into the Henderson-Hasselbalch equation:\[ 5.8 = 4.8 + \log\left( \frac{[\text{salt}]}{[\text{acid}]} \right) \]
03

Isolate the Logarithmic Term

Subtract 4.8 from both sides to isolate the logarithmic term:\[ 5.8 - 4.8 = \log\left( \frac{[\text{salt}]}{[\text{acid}]} \right) \]\[ 1.0 = \log\left( \frac{[\text{salt}]}{[\text{acid}]} \right) \]
04

Solve for the Ratio

To eliminate the logarithm, rewrite the equation using the property of logarithms that states if \( \log\, x = y \), then \( x = 10^y \):\[ \frac{[\text{salt}]}{[\text{acid}]} = 10^{1.0} \]\[ \frac{[\text{salt}]}{[\text{acid}]} = 10 \]
05

Calculate the Inverse Ratio

The question asks for the ratio \( \frac{[\text{acid}]}{[\text{salt}]} \). Take the inverse of the ratio found in Step 4:\[ \frac{[\text{acid}]}{[\text{salt}]} = \frac{1}{10} \] \[ \frac{[\text{acid}]}{[\text{salt}]} = 0.1 \]
06

Choose the Correct Answer

Compare the calculated ratio \( \frac{[\text{acid}]}{[\text{salt}]} = 0.1 \) with the choices provided:1. \(0.1\)2. 103. 14. 2The correct answer is (a) 0.1.

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.

Buffer Solution
A buffer solution is a special type of chemical mixture that resists changes in pH when acids or bases are added to it. This makes buffer solutions incredibly useful in maintaining stable conditions in various chemical reactions and biological processes. The buffer typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. These components work together to neutralize added acid or base.

In our discussion, we're dealing with a buffer solution made from a weak acid and its salt, which is the salt of the conjugate base. When acid is added to the buffer, the conjugate base will react to neutralize it. Conversely, when a base is added, the weak acid in the buffer will react to neutralize the base. This ability to maintain a constant pH is why buffers are so important in chemistry, especially in biochemistry where enzymes require specific pH levels to function effectively.
pH Calculation
The calculation of pH in a buffer solution can be accomplished using the Henderson-Hasselbalch equation. This equation is a simplified formula that helps estimate the pH of a buffer solution, defined as:
  • \[\text{pH} = \text{pKa} + \log\left( \frac{[\text{salt}]}{[\text{acid}]} \right)\]

This equation shows how the pH of a buffer is related to the pKa of the acid (a measure of its strength) and the ratio of the concentrations of the conjugate base (the salt) to the acid. In the exercise, the values given are pH = 5.8 and pKa = 4.8. When these are substituted into the equation, we solve the resulting expression to find the necessary ratio of the concentrations of salt to acid.
The logarithmic component of the equation plays an integral role in determining how changes in the ratio of salt to acid can influence overall pH levels.
Weak Acid
Weak acids are acids that do not dissociate completely in solution. This incomplete ionization is contrasted with strong acids, which almost entirely dissociate. The strength of a weak acid is represented by its pKa value—the lower the pKa, the stronger the acid.
  • In our scenario, the weak acid has a pKa value of 4.8, which gives us insight into its dissociation tendency and buffering capacity.
  • The Henderson-Hasselbalch equation utilizes the pKa to connect the dissociation of the weak acid to the pH of the solution, offering a practical approach to predict pH changes.

Understanding weak acids is crucial for predicting the behavior of buffer solutions, especially since it determines how well a buffer can actually function. These principles are applied in our exercise to determine the ratio of acid to salt for achieving a desired buffer pH.

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

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

In the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(0.5 \mathrm{~mol} \mathrm{~L}^{-1^{2}}\) and concentration of \(\mathrm{H}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) while concentration of \(\mathrm{S}_{2}\) is \(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\) in one litre vessel. The value of equilibrium constant of the reaction is: (a) \(0.016\) (b) \(0.013\) (c) \(0.020\) (d) \(0.030\)

The exothermic formation of \(\mathrm{ClF}_{3}\) is represented by the equation: \(\mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-329 \mathrm{~kJ}\). Which of the following will increase the quantity of \(\mathrm{CIF}_{3}\) in an equilibrium mixture of \(\mathrm{Cl}_{2}, \mathrm{~F}_{2}\) and \(\mathrm{ClF}_{3}\) ? (a) Increasing the temperature (b) Removing \(\mathrm{Cl}_{2}\)

The equilibrium constant for the reaction: \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}\). The value of \(\mathrm{K}_{c}\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4\. What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

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