# 10.0 mL of 0.121 M H2SO4 is nuetralized by 17.1 mL of KOH solution. The molarity of the KOH solution is?

The reaction between H2SO4 and KOH can be written as,

H2SO4 + 2KOH ---------------> K2SO4 + 2H2O

Therefore the stoichiometry between H2SO4 and KOH is,

H2SO4:KOH = 1:2

One mol of H2SO4 reacts with 2 mols of KOH.

Number of mols present in 10 ml of 0.121 M H2SO4 solution

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The reaction between H2SO4 and KOH can be written as,

H2SO4 + 2KOH ---------------> K2SO4 + 2H2O

Therefore the stoichiometry between H2SO4 and KOH is,

H2SO4:KOH = 1:2

One mol of H2SO4 reacts with 2 mols of KOH.

Number of mols present in 10 ml of 0.121 M H2SO4 solution

= (0.121/1000) * 10

= 0.00121 mol.

Therefore number of KOH mols required is,

= 2 x 0.00121 mol

= 0.00242 mol.

This is present in 17.1 ml of KOH solution, let's say its molarity is M, then,

0.00242 = (M/1000)* 17.1

M = 0.1415 M

The molarity of KOH solution is 0.1415 M.

Approved by eNotes Editorial Team