A volume of 70.0mL of aqueous potassium hydroxide(KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the solution if 19.2mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4->K2SO4+2H20
moles H2SO4 = M x L = ??
moles KOH = 2 x moles H2SO4.
M KOH = moles KOH/L KOH
posted by DrBob222
vai se fuder ---- > vai tomar no cu