Determine the volume of 0.230 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is:
H2SO4(aq)+2KOH(aq)→ K2SO4(aq)+2H2O(l)
Part A
25 mL of 0.230 M H2SO4. Express your answer using two significant figures.
We have this balanced equation here,
H2SO4 + 2KOH = K2SO4 + 2H2O
SO, we need 2 mole of KOH to neutralize 1 mole of H2SO4.
Simple solution we have,
(M1 × V1 )/ N1 =( M2 × V2) / N2
where,
M1 = molarity of H2SO4 = 0.230M
V1 = Volume of H2SO4 = 25 ml
N1 = no. Of mole of H2SO4 = 1
M2 = molarity of KOH = 0.230 M
V2 = volume of KOH = (?)
N2 = no. Of mole of KOH = 2
So,
(0.230 × 25)/1 = ( 0.230 × V2)/ 2
5.75 = V2 × 0.115
V2 = 50 ml.
we require 50 ml of 0.230 m KOH to neutralize 25 ml of 0.230 m H2SO4.
Here, notice that we need double volume of KOH.why?
because 2 mole of KOH only neutralize one mole of H2SO4.
And concentration are also same so we require 50 ml (25×2) i.e, double volume.
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Determine the volume of 0.230 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is:
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