A volume of 60.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.2 mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4--K2SO4+2H2O. I cant seem to get the right answer so please, if you know how to do this explain it fully and work out the problem completely with numbers and equations.

Question

A volume of 60.0  of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.2 mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4-->K2SO4+2H2O. I cant seem to get the right answer so please, if you know how to do this explain it fully and work out the problem completely with numbers and equations.

DrBob222 · Answer

Has it occurred to you that if you had posted your work I could have found the error in half the time it takes to work the problem.
H2SO4 + 2KOH ==> K2SO4 + 2H2O

How many moles H2SO4 were used? That is M x L = moles H2SO4.
Now convert that to moles KOH. From the equation, it takes 2 moles KOH to equal 1 mole H2SO4; therefore, moles H2SO4 x 1/2 = moles KOH.
Now M KOH = moles KOH/L KOH.