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 hydrogen ion concentration in a solution of weak acid of dissociation constant \(K_{a}\) and concentration \(C\) is nearly equal to (1) \(\sqrt{\frac{K_{\mathrm{u}}}{C}}\) (2) \(\frac{C}{K_{\mathrm{a}}}\) (3) \(K_{\mathrm{a}} \cdot C\) (4) \(\sqrt{K_{\mathrm{a}} \cdot \mathrm{C}}\)

Short Answer

Expert verified
The hydrogen ion concentration is \sqrt{K_a \cdot C}\, corresponding to option (4).

Step by step solution

01

Understand the Given Problem

The problem requires determining the hydrogen ion concentration in a solution of a weak acid. The given parameters are the dissociation constant \(K_a\) and the concentration of the acid \(C\).
02

Write the Dissociation Equation

For a weak acid \(HA\) dissociating in water, the reaction is: \[HA \rightleftharpoons H^+ + A^-\]
03

Write the Expression for the Equilibrium Constant

The equilibrium constant expression is: \[K_a = \frac{[H^+][A^-]}{[HA]}\]
04

Assume Degree of Dissociation is Small

Since it is a weak acid, the degree of dissociation is small. Hence, the concentration of \(HA\) remains nearly equal to the initial concentration \(C\).
05

Simplify the Expression

Assuming \([H^+] = [A^-]\), the expression simplifies to: \[K_a = \frac{[H^+]^2}{[HA]} \approx \frac{[H^+]^2}{C}\]
06

Solve for Hydrogen Ion Concentration

Rearranging the above equation, we get: \[ [H^+] = \sqrt{K_a C} \]
07

Compare with Given Choices

Comparing \([H^+] = \sqrt{K_a C} \) with the given options, the correct option is (4) \(\sqrt{K_a \cdot C}\).

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.

Weak Acid Dissociation
In chemistry, a weak acid only partially dissociates in water. When a weak acid such as HA is added to water, it splits into hydrogen ions (H+) and its conjugate base (A-). However, unlike strong acids, weak acids do not ionize completely. Most of the acid molecules remain undissociated in the solution.
For a weak acid dissociation, the reaction can be represented as:
\[HA \rightleftharpoons H^+ + A^-\]
Weak acid dissociation is influenced by several factors including the strength of the acid, the initial concentration, and the temperature. The degree of dissociation, which is the fraction of the original acid that has ionized, is typically small for weak acids. Understanding this concept is crucial when calculating the pH of weak acid solutions because it affects the concentration of hydrogen ions in the solution. If you assume the degree of dissociation is small, the acid's concentration doesn't significantly decrease from its initial value.
Equilibrium Constant
The concept of the equilibrium constant (K) is fundamental in chemistry. It helps predict the extent to which a reaction will proceed. For weak acids, the equilibrium constant is referred to as the acid dissociation constant (Ka).
In the case of a weak acid HA dissociating into H+ and A-, the equilibrium constant expression is:
\[K_a = \frac{[H^+][A^-]}{[HA]}\]
This equation basically states that at equilibrium, the product of the concentrations of the dissociation products divided by the concentration of the undissociated acid is equal to a constant value, Ka. This constant is unique to each acid and is a measure of its strength - a higher Ka value indicates a stronger acid that dissociates more in water.
In practical terms, for a weak acid with a small degree of dissociation, we assume that the change in concentration of the acid (C) is negligible. Therefore, the equilibrium concentrations of [HA] almost remain the same as the initial concentration.
Acid Dissociation Constant
The acid dissociation constant, denoted as Ka, is a specific type of equilibrium constant that applies to the dissociation of weak acids in aqueous solution.
Mathematically, it is expressed as:
\[K_a = \frac{[H^+][A^-]}{[HA]}\]
This constant provides a measure of the degree of ionization of the acid in water. Strong acids have large Ka values because they ionize completely, whereas weak acids have smaller Ka values due to their partial ionization.
To find the hydrogen ion concentration \([H^+]\) in a weak acid solution, you can use the Ka expression. When you set \([H^+] = [A^-]\), the equation simplifies to:
\[K_a = \frac{[H^+]^2}{C}\]
Rearranging this equation for \([H^+]\) gives:
\[ [H^+] = \sqrt{K_a \times C}\ \]
This formula is crucial for calculating the pH of weak acid solutions because pH is directly related to the hydrogen ion concentration. By taking the negative logarithm of \([H^+]\), you can find the pH, further providing practical insights into the acidity of the solution.

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 pII of a solution is \(5.0 .\) To this solution sufficicnt acid is added to decrease the \(\mathrm{pH}\) to \(2.0 .\) The increase in hydrogen ion concentration is (1) 1000 times (2) \(5 / 2\) times (3) 100 times (4) 5 times

Which of the following is correct? 1) \(K_{\mathrm{a}}\) (weak acid) \(\times K_{\mathrm{b}}\) (conjugate weak base) \(=K_{\mathrm{v}}\) 2) \(K_{\mathrm{a}}\) (strong acid) \(\times K_{\mathrm{b}}\) (conjugate strong base) \(=K_{\mathrm{w}}\) 3) \(K_{\mathrm{a}}\) (weak acid) \(\times K_{\mathrm{b}}\) (weak base) \(=K_{\mathrm{T}}\) 4) \(K_{\mathrm{a}}\) (weak acid) \(\times K_{\mathrm{b}}\) (conjugate strong base) \(=K_{\mathrm{w}}\)

Which of the following statements is falsc? (1) Hydrolysis of the salt of strong acid and weak base is called cationic hydrolysis (2) Hydrolysis of the salt of weak acid and strong base is known as anionic hydrolysis (3) Aqueous solution of aluminium chloride is acidic due to hydrolysis of \(\mathrm{Al}^{3}\) ion (4) Aqueous solution of sodium carbonate is basic due to hydrolysis of \(\mathrm{Na}\) ion

The correct statement is (1) \(\mathrm{NII}_{4} \mathrm{Cl}\) gives an alkalinc solution in water (2) \(\mathrm{CII}_{3}\) COONa gives an acidic solution in water (3) \(\mathrm{CII}_{3} \mathrm{COOII}\) is a weak acid (4) \(\mathrm{NII}_{4} \mathrm{OII}\) is a strong base

Which of the following mixture can act as a buffer solution? (1) I litre of \(0.2 \mathrm{~N} \mathrm{CH}_{3} \mathrm{COOH}+1\) litre of \(0.2 \mathrm{~N} \mathrm{NaOH}\) (2) I litre of \(0.1 \mathrm{~N} \mathrm{CH}_{3} \mathrm{COOH}+1\) litre of \(0.2 \mathrm{~N} \mathrm{NaOH}\) (3) 1 litre of \(0.2 \mathrm{~N} \mathrm{CH}_{3} \mathrm{COOH}+1\) litre of \(0.1 \mathrm{~N} \mathrm{NaOH}\) (4) All the above

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