To determine the correct sequence of acidic strength for the given aliphatic carboxylic acids, let's analyze each compound:
The electron-donating effect of alkyl groups decreases the acidity because they destabilize the carboxylate ion formed after deprotonation. Thus, the more extended the alkyl chain, the weaker the acid.
Based on the analysis above, the sequence of acidic strength in decreasing order is:
Therefore, the correct answer is: \(HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH\).
The acidic strength of carboxylic acids is influenced by the electron-withdrawing or electron-donating effects of alkyl groups:
HCOOH (formic acid) is the most acidic as it has no electron-donating alkyl group, which would reduce acidity.
CH$_3$COOH (acetic acid) is less acidic because the methyl group (CH$_3$) is weakly electron-donating.
CH$_3$CH$_2$COOH (propionic acid) is even less acidic due to the larger electron-donating ethyl group.
CH$_3$CH$_2$CH$_2$COOH (butyric acid) is the least acidic because the longer alkyl chain has a stronger electron-donating effect.
The correct order of acidic strength is:
\[ \textbf{HCOOH > CH$_3$COOH > CH$_3$CH$_2$COOH > CH$_3$CH$_2$CH$_2$COOH}. \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)