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A volume of 40.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What w...Asked by Sarah
A volume of 60.0mL of aqueous potassium hydroxide (KOH ) was titrated against a standard solution of sulfuric acid (H 2 SO 4 ). What was the molarity of the KOH solution if 25.7mL of 1.50 M H 2 SO 4 was needed? The equation is
2KOH(aq)+H 2 SO 4 (aq)¨K 2 SO 4 (aq)+2H 2 O(l)
2KOH(aq)+H 2 SO 4 (aq)¨K 2 SO 4 (aq)+2H 2 O(l)
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Answered by
bobpursley
you need twice the moles of KOH as H2SO4
molesKOH=2*molesH2SO4
molaritiy*.0257L=2*1.50*.0257
solve for molarity.
molesKOH=2*molesH2SO4
molaritiy*.0257L=2*1.50*.0257
solve for molarity.
Answered by
jonathan
should be 3 right?
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