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 Attempt at a Solution
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?