[BaO]

What is the pH of the solution formed when 0.040 mol KOH is added to 2.00 L of 0.0020 M HCl?
 
 this is how i did:
         [OH^-] _diL=[KOH]=0.020 M
         [H₃O⁺]_diL=[HCl]=0.0020 M
     
=> [OH^-]_excess=0.020 - 0.0020=0.018 M (because [OH^-]_used= [H₃O⁺]_diL)

      pH =14.00 - pOH=12.26

i keep getting that number while the answer is 7.00. could you help me out?

thanks a lot

[Albert]

Don't use molarities to solve it but moles.

[BaO]

you meant like this:
 moles of HCl = 0.004 moles = moles of  H₃O⁺

 moles of KOH = moles of OH^-=0.04 moles

 moles of OH^-_excess= 0.04 -0.004=0.036 moles

[OH^-]= 0.018 ( isnt it the same? or i misunderstand you?)

[lemonoman]

Both moles and molarity can be used in this situation...many people/teachers prefer using moles, because it's easier to picture one H+ and one OH- neutralizing each other, than it is to picture 0.1 M of H+ and 0.1 M OH- neutralizing each other.

That's why you get the same answer either way (only applicable for strong acids and strong bases...weak acid or weak base would make things SLIGHTLY more complicated...but not too much)

Anyways, what you've done is correct in my mind.  It seems highly likely to me, that there was a mistake in the question - not necessarily from you, bao....from the textbook/wherever you got it from.

Hope that helps

[Albert]

What is the pH of the solution formed when 0.040 mol KOH is added to 2.00 L of 0.0020 M HCl?

I think the mistake is in molarity of HCl: maybe there are too many zeros.

However, you got the right result, considering the data you posted.

[sdekivit]

if pH must be 7,00 i think you meant [HCl] = 0,02 M instead of 0,002 M