# Need to figure out this problem :(

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A 25.00-mL sample of H2SO4 solution of unknown concentration requires 28.27 mL of a 0.185 M KOH solution to complete the neutralization reaction. What is the concentration of the unknown H2SO4 solution?

### H2SO4 (aq) + 2 KOH (aq) --

H2SO4 (aq) + 2 KOH (aq) ---> K2SO4 (aq) + 2 H2O (l)
Here is the balanced reaction between H2SO4 and KOH , the mole- mole ratio of H2SO4 and KOH is 1:2 , and in the given problem you can find moles of KOH with volume and molarity
Molarity = moles of solute / volume in liters
Convert given volume in ml to l .02827 L and
.185 = moles / 0.02827
Moles = y moles of KOH
now by using stoichiometry you can find the moles of H2SO4 and then concentratin using its given volume
y moles KOH X 1 mole H2SO4 / 2 mole KOH = z moles of H2SO4
And then convert volume of H2SO4 in ml to l
And use concentration= moles of KOH/ volume of solution in Liters

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