(adsbygoogle = window.adsbygoogle || []).push({}); A solution of KOH is prepared by dissolving 2.00g of KOH in water to a final volume of 250mL of solution. What volume of this solution will neutralize 20.0mL of 0.115 mol/L sulfuric acid?

2. Relevant equations

C=n/v

C1V1=C2V2

n=m/M

Attempt at solution

First I balanced the equation:

2KOH+H2SO4->k2S04+2H20

Than I figured out the moles of H2S04:

v=0.02L

C=0.115mol/L

n=c*V=2.3x10^-3 mols.

Than I found the mols of KOH

m=2g

M=56.1g/mol

n=m/M.

n=0.0356 mols

The problem is after this step I get confused as to what to do with the 250mL. Can someone give me an explanation please on how and why? (if you help me solve it)

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