Calcutating the volume of HNO3

  • Thread starter h20h
  • Start date
  • #1
18
0

Homework Statement



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






Homework Equations


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?
 

Answers and Replies

  • #2
symbolipoint
Homework Helper
Education Advisor
Gold Member
6,324
1,241
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
18
0
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
 

Related Threads on Calcutating the volume of HNO3

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
5
Views
10K
Replies
4
Views
12K
Replies
3
Views
11K
  • Last Post
Replies
5
Views
2K
Replies
2
Views
5K
Replies
9
Views
964
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
5
Views
1K
Top