This equation shows that the acid strength of each of these hydrogen halides is a measure of the ease of ionization of the H—X bond (X = a halogen). The stronger the bond, the weaker the acid. Understanding how to determine which H— bond is stronger will allow you to predict the relative strength of each acid.
Determining which H—X bond is stronger requires
consideration of the differences among the various halogens. Two significant differences between fluorine (F), the element at the top of the periodic table in the halogen family, and iodine (I), an element near the bottom are: F is more electronegative than I thus allowing F to hold its electrons more tightly. Also F is much smaller than I because F has fewer electrons. Being larger, I is more polarizable and therefore softer than F. The higher electronegativity of F would seem
to result in a more stable ion, whereas, the softness of I means that the bond between I and the hard H is weaker than the bond between the hard F and the hard H. These predictions are obviously true as shown by the pKa of the two acids: HF has a pKa of 3.2, whereas HI ismuch stronger with a pKa of –10.
Because a Brønsted-Lowry acid donates a proton (H⊕) to a base, the point of comparison for the relative strengths of the various acids is the atom, or group of atoms, to which the proton is attached. This atom or group of atoms bears the charge in the conjugate base. The more stable the conjugate base, the higher the acid strength of the Brønsted-Lowry acid. For example, HF with a pKa of 3.2 is a much stronger acid than CH4 with a pKa of 49. This occurs because the F is
more electronegative than C so F holds its electrons more tightly than C. Thus, the F anion is a weaker base than a C anion. Because F is a weaker base, it can donate a proton more readily than can C. Within a period on the periodic table, Brønsted-Lowry acidity increases from left to right: for example, CH4 < NH3 < H2O < HF.