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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: h20h on April 14, 2007, 12:59:21 AM

Title: Calculating volume of HNO3 necessary to neutralize?
Post by: h20h on April 14, 2007, 12:59:21 AM

Q:  Calculate the volume of 0.6M HNO3 necessary to neutralize 28.55 mL of 0.45M KOH?

My solution:  HNO3 + KOH = H20 + KNO3    (KNO3 is the neutral salt)
the stoich is 1:1:1:1, so I took the 28.55 mL of KOH and converted to L which came out to be .02855 liters of KOH.  I then took those liters and multiplied by .45M and divided by 1 liter to get .0128 moles KOH.  Since the stoich is 1:1  the moles of KOH is the same as the moles of HNO3.  So I took the 0.128 moles of HNO3 and divided by the .6M that was given in the question..since M equals # of moles/liter of solution I just rearranged that equation to solve for liters and the result is .0213 liters and then multiplied by 1000 to get 21.3mL?   

Any suggestions, help here?

Thanks

Title: Re: Calculating volume of HNO3 necessary to neutralize?
Post by: Borek on April 14, 2007, 12:05:05 PM

I got 21.41 mL.

Title: Re: Calculating volume of HNO3 necessary to neutralize?
Post by: h20h on April 14, 2007, 12:10:47 PM

Did I do something wrong in the calculations? 

This is what I have:

28.55mL/1000mL  =  .02855 liters(.45M)/1liter =  .0128 moles of KOH

stoich 1:1

.0128 moles of HN03/.6M  =  .0123 liters(1000ml)  21.3 mL needed

am I off somewhere???

Thanks and let me know

Title: Re: Calculating volume of HNO3 necessary to neutralize?
Post by: Borek on April 14, 2007, 02:21:46 PM

Don't round off intermediate results.