Question

$H_{2}S{O}_4$ is a strong acid than $H_{2}S{O}_3$. If true enter 1, else 0.

Answer 1

Hint: Higher the oxidation number of the central atom, higher would be the acidity of oxyacids. Thus by finding the oxidation state of Sulphur in $H_{2}S{O}_4$ and $H_{2}S{O}_3$ ,we will get to know which acid is more strong.

Complete step by step solution:
-As we know the term acid and base have been defined in different ways in chemistry. Generally, we can say that an acid is any hydrogen-containing substance which is capable of donating hydrogen ion(proton) to another substance.
- According to Arrhenius, acids are compounds which ionize to produce hydrogen ions and according to Lowry-Bronsted definition, an acid is a proton donor. By the Lewis definition, acids are molecules or ions capable of coordinating with unshared pairs of electrons or in other words to be acidic a molecule must be deficient of electron pairs.
 -Solutions can be classified as acidic or basic on their hydrogen ion concentration relative to pure water. If the pH below 7 then the solution is acidic and if it's greater than seven the solution is said to be basic.
- For oxyacids, we can decide the strength of the acids according to the oxidation state of the central atom. The greater the oxidation state, greater would be the acidic strength. As we mentioned, as the oxidation state of the central atom becomes larger, the acidity of the molecule increases.
- An atom will become more electronegative as the oxidation number increases. This is because the OH bond will become more polar and therefore more acidic. Since, acidic strength is directly proportional to electronegativity, higher the oxidation state, higher would be the acidity.
- The oxidation state of Sulphur in $H_{2}S{O}_4$ is $+$6 and in $H_{2}S{O}_3$ the oxidation state is $+$4.
- Hence $H_{2}S{O}_4$ is more electronegative than $H_{2}S{O}_3$ .

Therefore, the answer is $H_{2}S{O}_4$ is a strong acid than $H_{2}S{O}_3$.

Note: We can check the strength of acid also by looking at how stable its conjugate base is. The more the number of resonating structures the base can form, the higher would be the stability of the base. As we can calculate, Sulphate ion ($S{O_4}^{2-}$) has 6 resonating structures which makes it a more stable base than Sulphite ion ($S{O_3}^{2-}$) which has only 3 resonating structure. Thus $H_{2}S{O}_4$ would be more acidic.