If 31.3 mL KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution

Can you please help me understand how to find the answer to these two problems? I need to understand how to solve them in order to be ready for my exam on Wednesday :(

I need help understanding these two Chemistry problems:

1)
A solution of 0.304 M KOH is used to neutralize 19.0 mL H3PO4 solution.
H3PO4 (aq) + 3KOH(aq) ----> K3PO4 (aq) + 3H2O (l)

----> If 31.3 mL KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution?

2)
A solution of 0.162 M NaOH is used to neutralize 25.0 mL H2SO4 solution.

H2SO4 (aq) + 2NaOH (aq) -----> Na2SO4 (aq) + 2H2O (l)

------> If 32.4 mL NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?


Thank you so much for your much appreciated help!! I really appreciate it! :)

#1
Take first reactant = KOH ( M1 , V1 , n1 are its respective molarity , volume , moles involved )

Second reactant = H3PO4 ( M2 , V2 , n2 are its respective molarity , volume , moles involved )


Let M1 = 0.304

V1 = 31.3 mL = 0.0313 L

n1 = 3


Let molarity of H3PO4 reqd = M2

V2 = 19 mL = 0.019 L

n2 = 1


Neutralization-Molarity relation is

M1V1 / n1 = M2V2 / n2


0.304 x 0.0313 / 3 = M2 x 0.019 / 1

Solving for M2


M2 = 0.1669 M

Molarity of H3PO4 = 0.1669








2) Similarly you can take

M1 , V1 , n1 for first reactant i.e H2SO4

M2 , V2 , n2 for second reactant i.e NaOH


Let molarity of H2SO4 reqd = M1

V1 = 25 mL = 0.025 L

n1 = 1

M2 = 0.162 M

V1 = 32.4 mL = 0.0324 L

n2 = 2



Applying the same relation

M1V1 / n1 = M2V2 / n2



M1 x 0.025 / 1 = 0.162 x 0.0324 / 2

Solving for M1

M1 = 0.104976 M


Molarity of H2SO4 = 0.104976 M




:- Remember

Convert the volume into L first

Apply simply the molarity - neutralization relation i.e

M1V1 / n1 = M2V2 / n2