write the balanced molecular equation for the neutralization reaction between H2SO4 and KOH in aqueous solution

Question

phases are optional

Answer 1

H₂SO₄ + KOH

The H₂SO₄ is a di-protic acid, i.e. it has 2 hydrogens (H⁺) that can react with the OH^- from KOH.  So, it stands to reason that it will take 2 KOH for each 1 H₂SO₄.  Right?

The balanced molecular equation would thus be H₂SO₄ + 2KOH ==> 2H₂O + K₂SO₄

All phases are (aq) except H₂O which is (l).

Answer 2

This is a neutralization reaction and also a double displacement reaction. Therefore, we can predict the products by writing each of our reactants out as the constituent ions that exist when they dissociate in aqueous solution and then switching the anion and cation partners:

H₂SO₄ dissociates into H⁺ and SO_4^2-

KOH dissociates into K⁺ and OH^-

 

When the cations and ions in solution switch partners, H+ joins with OH- to form water (H₂O), and K+ joins with SO₄^2- to for K₂SO₄.

 

So all together, the molecular equation is

 H₂SO₄ + KOH --> H₂O + K₂SO₄

 

When balanced, it is 

H₂SO₄ + 2KOH --> 2H₂O + K₂SO₄