Factors Affecting Acid Strength

What Makes One Acid Stronger Than Another?

Having learnt to calculate the pH of acid solutions, a natural question follows: why does one acid donate its proton more readily than another? The extent of dissociation of an acid HA depends on two properties of the H–A bond:

1. The strength of the H–A bond. The weaker the bond, the more easily it breaks, and the stronger the acid.
2. The polarity of the H–A bond. The greater the electronegativity difference between H and A, the more charge separation the bond already carries ($\text{H}^{\delta+}\text{—A}^{\delta-}$), the easier heterolytic cleavage becomes — and the stronger the acid.

Which of the two factors dominates depends on whether you compare elements down a group or across a period.

Comparing Down a Group: Bond Strength Decides

When the elements A belong to the same group of the periodic table, the size of A increases sharply down the group. The H–A bond becomes longer and much weaker, and this effect overwhelms the change in polarity.

For the hydrogen halides (Group 17), acid strength increases down the group:

Even though H–F is the most polar of these bonds (fluorine is the most electronegative element), it is also by far the strongest bond — so HF is only a weak acid in water, while HI, with its long, weak bond, is one of the strongest known acids. The same logic makes $\text{H}_2\text{S}$ a stronger acid than $\text{H}_2\text{O}$: the S–H bond is weaker than the O–H bond.

Comparing Across a Period: Polarity Decides

When the elements A sit in the same period, their sizes — and hence the H–A bond strengths — are similar. Now the polarity of the H–A bond becomes the deciding factor: as the electronegativity of A increases from left to right, the bond becomes more polar and the acid becomes stronger.

For the period-2 hydrides:

$\text{CH}_4 < \text{NH}_3 < \text{H}_2\text{O} < \text{HF}$ …