M1V1n1 = M2V2n2 Suppose that 17.38 mL of 0.02765 M aqueous H2SO4 is required to neutralize 10 mL of an aqueous solution of KOH.?

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

H₂SO₄: M₁ = 0.02765 M, V₁ = 17.38, n₁ = 2
KOH: M₂ = ? M, V₂ = 10 mL, n₂ = 1

M₁V₁n₁ = M₂V₂n₂
(0.02765 M) × 17.38 × 2 = M₂ × 10 × 1

Molarity of KOH = 0.02765 M × 17.38 × 2 / 10 M = 0.09611 M

Find the molarity of KOH.....

Simply use the unit-factor method as you would for any stoichiometry problem. The M1V1=M2V2 equation is for dilutions and has been co-opted to solve titration problems.

Assume that 10 mL is really 10.00 mL in order to maintain the four significant digits of the other measurements.

H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2HOH(l)
17.38 mL ....... 10.00 mL
0.02765M ........ ???M

0.01738L x (0.02765 mol H2SO4 / 1L) x (2 mol NaOH / 1 mol H2SO4) / 0.01000L = 0.09611M NaOH