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

Question

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

DrBob222 · Answer

moles H2SO4 = M x L = ?? moles KOH = 2 x moles H2SO4. M KOH = moles KOH/L KOH

Anonymous · Answer

vai se fuder ---- > vai tomar no cu

Anonymous · Answer

12.8