chemistry
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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Answers
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The neutralization reaction between KOH and H2SO4 (sulfuric acid) can be written as:
KOH + H2SO4 → KHSO4 + H2O
From the given information, the concentration of KOH (mol/L) = 0.357mol/25ml = 14mol/L
The amount of sulfuric acid (H2SO4) is 0.115mol/20ml = 0.0058mol
The amount of KOH required to neutralize 0.0058mol of sulfuric acid would be equal to 0.0058mol.
Therefore, the volume of the KOH solution required to neutralize 20.0ml of 0.115mol/L sulfuric acid is
Volume of KOH (ml) = 0.0058mol/14mol/L = 0.0414ml
