# Calcutating the volume of HNO3

1. Apr 14, 2007

### h20h

1. The problem statement, all variables and given/known data

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

2. Relevant equations
My solution: HNO3 + KOH = H20 + KNO3 (KNO3 is the neutral salt

3. 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?

2. Apr 14, 2007

### symbolipoint

Are you saying "moles of nitric acid equals moles of potassium hydroxide"?
Are then equating the two expressions which tell those moles?

Ma*Va = Mb*Vb

The unknown value is the variable, Va. (excuse the notation here not using true subscripts)

3. Apr 14, 2007

### h20h

Yes the stoichiometry of the equation is 1:1 so the moles of KOH is equivalent to the moles of HNO3. I must just be plugging something in wrong ? can you tell by what I have written? I know you are getting 21.4 and I am 21.3 but still I want to be right

Thanks and let me know