Question

$$H_{2}S{O}_4$$ is a stronger acid than $H_{2}S{O}_3$ .
A. True
B. False

Answer 1

Hint: In order to the question, first we should compare the properties and composition of both of the compounds. And also write the properties by which acidity will expand and then compare with both compounds.

Complete answer:
Yes, $$H_{2}S{O}_4$$ ​is a stronger acid than $H_{2}S{O}_3$ .
The oxidation state of $S$ is more in $$H_{2}S{O}_4$$ which is +7 as compared to +4 in $H_{2}S{O}_3$ .
Greater is the oxidation state of the central atom, the more it attracts the electrons towards itself and also, more it stabilises the conjugate base.
At the point when two compounds have a similar overabundance oxygen count, the next factor to consider is the electronegativity of the central atom. The more electronegative the central atom, the more it pulls out electron density from the acidic proton, expanding the acidity. So, we can say that electronegativity increases as we go left to right in the periodic table and as the electronegativity increases, acid strength increases. That’s why Sulphuric acid or $$H_{2}S{O}_4$$ is more stronger acid than Sulfurous acid or $H_{2}S{O}_3$ . The key difference between sulphuric acid and sulfurous acid is in the oxidation number of Sulphur. In other words, sulphuric acid is a very strong acid, and sulfurous acid is relatively weak. The key difference between sulphuric acid and sulfurous acid is in the oxidation number of Sulphur.
Hence, the correct option is (A.) True .

Note:We can write the another reason such that Sulphuric acid is a diprotic acid, implying that it yields two protons $(H^{+})$ per atom when it dissociates in water. It is a strong acid in light of the fact that the disassociation of the main proton is highly preferred.