A solution of KOH is prepared by dissolving 2.00 g of KOH in water to a final volume of 250 ml of solution what volume of this solution will neutralize 20.0 ml of 0.115 mol/L sulfuric acid?
amount of KOH= 2.00g(1mol of KOH/ 56.01g)= 0.357 moles
concentration of KOH in water = amount of KOH/ volume of kOH= ( 0.357 mol/(25x10^-3)= 14 mol/ l
please help

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  1. molar mass KOH is 56.1 for starters and
    2.00/56.1 = 0.0357
    M KOH = 0.0357/0.250 = 0.143 instead of your number of 14.

    H2SO4 + 2KOH ==> K2SO4 + + 2H2O
    mols H2SO4 = M x L = ?
    Using the coefficients in the balanced equation, convert mols H2SO4 to mols KOH.
    Then M KOH = mols KOH/L KOH. YOu know M KOH and mols KOH, solve for L KOH

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