Saturday, April 24, 2010

Average moles of NaOH requred to neutralize the HCl?

i had a sub today and they didn't know how to help me on this question. and i can't find anywhere how to do this question. can you guys please help me ?





i'm given the information that Volume of 0.109 M NaOH requred to neutralize 25.0 mL of HCl.





One of the questions are: determine the average moles of NaOH required to neutralize the HCl.


i tried doing this question, but i dont know if i did it correct.. i got 0.190M = 0.190mol/L x 0.025L = 0.0475mol


i dont know if thats right though.. =/





also.. how do u determine how many moles of HCl were neutralized by the moles of NaOH used. i have NO idea how to do that question.





thanks in advance =)!!

Average moles of NaOH requred to neutralize the HCl?
it depends on how many moles of HCl you have, which depends on the molarity of the 25mL of HCl (i doubt youre using 25mL of liquified HCl)





HCl and NaOH neutralize on a 1:1 ratio (acid-base reaction)


you are neutralizing the acidic H from HCl with the basic OH from NaOH





after they react, you are left with H20 (water) and NaCl (table salt)





are you sure you wrote the question right?
Reply:You need the concentration of HCl, unless you have it you can not solve this question.


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