The basicity of tetraoxosulphate(vi) is 2. Explain this statement with an equation showing the reaction between the acid and potassium hydroxide.
The basicity of an acid is the number of H atoms replaced in a reaction with KOH.
2KOH + H2SO4 ==> K2SO4 + 2H2O
To understand the basicity of tetraoxosulphate(VI), we need to look at its chemical formula and its reaction with potassium hydroxide.
The chemical formula for tetraoxosulphate(VI), also known as sulphuric acid, is H2SO4. It is important to note that sulphuric acid is a strong acid, meaning it can fully ionize in water to produce hydrogen ions (H+) and sulphate ions (SO4^-2).
When sulphuric acid reacts with potassium hydroxide (KOH), a neutralization reaction occurs. The hydrogen ions from sulphuric acid combine with hydroxide ions (OH-) from potassium hydroxide to form water (H2O). Meanwhile, the sulphate ions combine with potassium ions (K+) to form potassium sulphate (K2SO4).
The balanced equation for this reaction is:
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the equation, we can see that for every one molecule of sulphuric acid, it reacts with two molecules of potassium hydroxide to form two molecules of water and one molecule of potassium sulphate. Therefore, the basicity of tetraoxosulphate(VI) is 2, as it requires two molecules of potassium hydroxide to neutralize one molecule of the acid.