you need twice the moles of KOH as H2SO4
molesKOH=2*molesH2SO4
molaritiy*.0257L=2*1.50*.0257
solve for molarity.
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)
2 answers
should be 3 right?
1 answer