If 72.1 ml of 0.543 M H2SO4 is needed to neutralized 39.0 ml of KOH solution, what is the molarity of the KOH solution? Show the balanced equation.
Thank you (will choose best answer)
Thank you (will choose best answer)
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2KOH+H2SO4--> K2SO4 +2H2O
you see , you need 2 molecules of KOh to neutralize 1 mole of H2SO4
we calculate the number of moles of H2SO4 in 72.1mL=0.072L
1liter of 0.543M of H2SO4 contains 0.543*0.072=0.0391moles of H2SO4
as you need 2 moles of KOH to neutralize 1mole of H2SO4 you need 0.0391*2=0.0782moles of KOH
which are in 39.0 mL = 0.039L
So the molarity of your solution is 0.0782/0.039=2.004M or 2M rounded
you see , you need 2 molecules of KOh to neutralize 1 mole of H2SO4
we calculate the number of moles of H2SO4 in 72.1mL=0.072L
1liter of 0.543M of H2SO4 contains 0.543*0.072=0.0391moles of H2SO4
as you need 2 moles of KOH to neutralize 1mole of H2SO4 you need 0.0391*2=0.0782moles of KOH
which are in 39.0 mL = 0.039L
So the molarity of your solution is 0.0782/0.039=2.004M or 2M rounded