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titration quick question

A student titrates a standard solution of barium hydroxide, Ba(OH)2, with nitric acid, HNO3.
25.00cm3 of 0.0450moldm–3 Ba(OH)2 are needed to neutralise 23.35cm3 of HNO3(aq). What is the concentration, in mol dm–3, of the nitric acid?

I thought the answer would be 0.0241 but its actually 0.0964, I worked out the moles of barium hydroxide and divided this by two and then divided that by the volume of nitric acid.
How do i get an answer of 0.0964?
Reply 1
The problem probably is that one mole of Ba(OH)2 reacts with 2 moles of HNO3. So if 1.125x10-3 is moles of Ba(OH)2, then moles of HNO3 is 2.25x10-3

Ba(OH)2 + 2NHO3 —-> Ba(NO3)2 + 2H2O
Reply 2
Moles of Ba(OH)2
= 25 x 0.045 x 10^-3 = 1.125 x 10^-3
Moles of NHO3 = 2 x 1.125 x 10^-3 = 2.25 x 10^-3
Conc of HNO3
= (2.25 x 10^ -3) / (23.35 x 10^-3) = 0.0963597...... = 0.0964 mol dm^-3
Reply 3
Oh right that makes sense! Just didn't understand the moles bit but it makes sense from the equation.
Thank you!!
Original post by Deggs_14
Moles of Ba(OH)2
= 25 x 0.045 x 10^-3 = 1.125 x 10^-3
Moles of NHO3 = 2 x 1.125 x 10^-3 = 2.25 x 10^-3
Conc of HNO3
= (2.25 x 10^ -3) / (23.35 x 10^-3) = 0.0963597...... = 0.0964 mol dm^-3
Hey dude, I was doing the question today and got it wrong cause I initially did what you did. To get the correct answer of 0.0964, I basically did the following (I’m gonna refer to the concentration we are looking for as A):25/1000 x 0.045= 23.35/1000 x A Cancel out the 1000 on both sides to get this:25 x 0.045 = 23.35 x AThen rearrange to get A on it’s own: (25 x 0.045) / 23.35 This equals 0.04817 Now we know that in the equation, for one mole of Ba(OH)2, we have 2 moles of HNO3, so we multiply 0.04817 by 2. This gives us 0.09636, which rounded gives us 0.0964 :smile:. Good luck next week!
NOOOOO I DIDNT SEE THE REPLY U GOT 😭😭 imagine I was actually excited to reply to this 😣.
Original post by luckstar606
A student titrates a standard solution of barium hydroxide, Ba(OH)2, with nitric acid, HNO3.
25.00cm3 of 0.0450moldm–3 Ba(OH)2 are needed to neutralise 23.35cm3 of HNO3(aq). What is the concentration, in mol dm–3, of the nitric acid?

I thought the answer would be 0.0241 but its actually 0.0964, I worked out the moles of barium hydroxide and divided this by two and then divided that by the volume of nitric acid.
How do i get an answer of 0.0964?
Omg someone tell me whether this is a correct way to work it out cause I’ve been using this and have so far gotten right answers when I use this but like the above reply used something totally different????
Original post by afsay
Hey dude, I was doing the question today and got it wrong cause I initially did what you did. To get the correct answer of 0.0964, I basically did the following (I’m gonna refer to the concentration we are looking for as A):25/1000 x 0.045= 23.35/1000 x A Cancel out the 1000 on both sides to get this:25 x 0.045 = 23.35 x AThen rearrange to get A on it’s own: (25 x 0.045) / 23.35 This equals 0.04817 Now we know that in the equation, for one mole of Ba(OH)2, we have 2 moles of HNO3, so we multiply 0.04817 by 2. This gives us 0.09636, which rounded gives us 0.0964 :smile:. Good luck next week!
Reply 7
at least you got the right answer lool 'excited to reply to this' the joys of chemistry a level :///thank you for replying though!
Hey, I'm pretty late (2 years to be exact)- but how did you find out there were two moles of HNO3? And do we need to write out the equation for these kinds of questions?
Original post by Sq-ish-y
Hey, I'm pretty late (2 years to be exact)- but how did you find out there were two moles of HNO3? And do we need to write out the equation for these kinds of questions?

Sometimes you're given the equation, but AQA often ask you to do this yourself. Ba(OH)2 gives out two moles of OH- when it dissolves in water. We're going to need two moles of H+ ions to react with them to make two moles of water. 2H+ + 2OH- —-> 2H2O. HNO3 can only give out one mole of H+ ions, HNO3 —-> H+ + NO3- so we're going to need two moles of nitric acid.

Ba(OH)2 + 2HNO3 —-> Ba(NO3)2 + 2H2O
(edited 1 year ago)
Original post by tony_dolby
Sometimes you're given the equation, but AQA often ask you to do this yourself. Ba(OH)2 gives out two moles of OH- when it dissolves in water. We're going to need two moles of H+ ions to react with them to make two moles of water. 2H+ + 2OH- —-> 2H2O. HNO3 can only give out one mole of H+ ions, HNO3 —-> H+ + NO3- so we're going to need two moles of nitric acid.

Ba(OH)2 + 2HNO3 —-> Ba(NO3)2 + 2H2O

thank you so much for the reply! I wasn't expecting someone to reply after so long haha- but yes it makes sense now. So the formula is kind of like a ratio?
Original post by Sq-ish-y
thank you so much for the reply! I wasn't expecting someone to reply after so long haha- but yes it makes sense now. So the formula is kind of like a ratio?

Exactly. The thing to remember is never to change the small numbers in the bottom right hand corner. In other words, nitric acid is always HNO3. This is written in stone and cannot be changed. If you change this number you change what the compound actually is. SO2 is sulfur dioxide, but SO3 is sulfur trioxide.
Knowing that the formula of the nitrate ion is always NO3- helps you build up formulas for other compounds. Ionic compounds are always neutral beacuse the ions balance each other out. If the positive ion is 2+, you're going to need two minuses. You can do this by using one ion which itself is 2- (the O2- ion, for example) or by using two -1 ions (such as the Cl- ion).

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