Question

What be the molarity of a KOH solution if, during a titration with 0.50 M HCl, 34.5 mL of the HCl neutralized 22.4 mL of the KOH solution?

a.) How much of a .50 M solution of H2SO4 would be needed to neutralize the same amount of KOH solution?

Answer 1

The balanced equation for KOH and HCl is given as

HCl+KOH---->KCl +H₂O

In this equation the mole ratio of HCl and KOH is 1:1

The moles of the titrant and analyte are equal is known as equivalence point.It is given as M₁V₁=M₂V₂

Where M₁ is Molarity of acid , V₁ is volume of acid ,M₂ is Molarity of base , V₂ is volume of base.

M₁= 0.50M V₁=34.5ml M₂ =? V₂=22.4ml

Therefore M₂=(0.50×34.5)÷22.4 =0.77M

a) similarly the equation for H₂SO₄ and KOH is given as

H₂SO₄ +2KOH ---> K₂SO₄+2H₂O

In this equation the mole ratio of acid and base is 1:2

Therefore M_acid ×V_acid =2× M_base ×V_base . Given that same amount of KOH is used. Therefore volume of H2SO4 is,

V_acid=(2×0.77×22.4)÷0.50 =68.99ml .