Asked by Paris
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.
Answers
Answered by
DrBob222
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.
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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