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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?
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)
Homework 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)