Question

Arrange the following in increasing order of their $pK_a$ values.

(x) $CH_3-S-O-H$ (y) $CH_3 - \underset{O}{\underset{||}{C}} - O - H$ (z) $CH_3OH$

• A. y < x < z
• B. x < y < z
• C. y < z < x
• D. x < z < y

Answer 1

Answer: (B)

x < y < z

Answer 2

The acidity of a compound is inversely related to its $pK_a$ value. Stronger acids have lower $pK_a$ values. We need to compare the acidity of methanesulfonic acid ($CH_3SO_3H$), acetic acid ($CH_3COOH$), and methanol ($CH_3OH$).

1. Methanesulfonic acid ($CH_3SO_3H$): This is a sulfonic acid. The conjugate base ($CH_3SO_3^-$) is highly stabilized by resonance due to the presence of electronegative oxygen atoms and the sulfur atom in a high oxidation state. Sulfonic acids are strong acids.
2. Acetic acid ($CH_3COOH$): This is a carboxylic acid. The conjugate base ($CH_3COO^-$) is stabilized by resonance between the two oxygen atoms. Carboxylic acids are weaker acids than sulfonic acids but stronger than alcohols.
3. Methanol ($CH_3OH$): This is an alcohol. The conjugate base ($CH_3O^-$) is an alkoxide ion and is not stabilized by resonance. Alcohols are weak acids.

The order of acidity is: $CH_3SO_3H > CH_3COOH > CH_3OH$

Since $pK_a$ is inversely proportional to acidity, the order of increasing $pK_a$ is the reverse of the acidity order: $pK_a(CH_3SO_3H) < pK_a(CH_3COOH) < pK_a(CH_3OH)$

Let (x) be $CH_3SO_3H$, (y) be $CH_3COOH$, and (z) be $CH_3OH$. Therefore, the increasing order of $pK_a$ values is $x < y < z$.