(i)
Give a brief explaination of the two bonding factors which determine the strength of an acid.
(ii)
Give a brief explanation of an Acid Buffer.
(iii)
Define an acid and base in terms of the Bronsted-Lowry theory. Give two examples of conjugate acid/base pairs which adhere to this theory
(iv)
Calculate the pH of a 0.02 M solution of benzoic acid which has a Ka of 6.3 x 10-5 at 25 oC where
C6H5COOH C6H5COO- + H
Thank you :)
This looks like a test. I will be happy to critique your thinking.
On iv, figure the conc of H+ first.
Ka=[H][C5H5COO-]/.02
Ka= x^2/.02 where conc H+ is x.
then, pH= -log x
(i) The two bonding factors that determine the strength of an acid are the polarity of the bond between the acid hydrogen and the molecule it is attached to, and the stability of the conjugate base formed after the acid donates its hydrogen ion.
To determine the polarity of the bond, you can look at the electronegativity difference between the acid hydrogen and the atom it is attached to. If the electronegativity difference is large, the bond will be polar, making it easier for the acid to release its hydrogen ion, thus increasing its acidity.
The stability of the conjugate base is determined by the resonance and inductive effects within the molecule. Resonance occurs when electrons can move freely between different atoms, stabilizing the negative charge on the conjugate base. Inductive effects result from the electron-withdrawing or electron-donating groups attached to the molecule, affecting its stability. The more stable the conjugate base, the stronger the acid.
(ii) An acid buffer is a solution that resists changes in pH when small amounts of acid or base are added to it. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid.
When a small amount of acid is added to an acid buffer, it reacts with the conjugate base present in the buffer, forming the weak acid. This helps maintain the pH by preventing a drastic change in the concentration of free hydrogen ions. Similarly, when a small amount of base is added to an acid buffer, it reacts with the weak acid to form the conjugate base, helping to maintain pH stability.
(iii) According to the Bronsted-Lowry theory, an acid is a substance that can donate a proton (H+) and a base is a substance that can accept a proton (H+).
Examples of conjugate acid/base pairs that adhere to this theory are:
1. Acetic acid (CH3COOH) and its conjugate base, acetate ion (CH3COO-).
In the reaction: CH3COOH + H2O ↔ CH3COO- + H3O+
Acetic acid donates a proton (H+) to water, acting as an acid, while water accepts the proton, acting as a base.
2. Ammonia (NH3) and its conjugate acid, ammonium ion (NH4+).
In the reaction: NH3 + H2O ↔ NH4+ + OH-
Ammonia accepts a proton (H+) from water, acting as a base, while water donates the proton, acting as an acid.
(iv) To calculate the pH of a 0.02 M solution of benzoic acid (C6H5COOH) with a Ka of 6.3 x 10^-5 at 25 oC, we need to consider the concentration of H+ ions formed when benzoic acid dissociates.
The dissociation reaction of benzoic acid is: C6H5COOH ↔ C6H5COO- + H+
From the equation, we can see that the concentration of H+ ions formed is equal to the concentration of benzoic acid that dissociates.
Using the Ka expression: Ka = [C6H5COO-][H+] / [C6H5COOH]
Given that the initial concentration of benzoic acid is 0.02 M and the Ka is 6.3 x 10^-5, we can assume that a small fraction of benzoic acid dissociates. Therefore, we can neglect the change in concentration of [C6H5COOH] by approximate it as 0.02 M.
Now, rearranging the Ka expression: [H+] = (Ka x [C6H5COOH]) / [C6H5COO-]
[H+] = (6.3 x 10^-5) x (0.02) / 0.02
[H+] = 6.3 x 10^-5
Taking the logarithm (base 10) of [H+]: pH = -log10(6.3 x 10^-5)
Calculating this, we find that the pH of the solution is approximately 4.20.