Posted by **Paris** on Saturday, June 25, 2011 at 9:27pm.

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.

- Chemistry -
**DrBob222**, Saturday, June 25, 2011 at 10:18pm
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.

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