Calculate moles of remaining in acid + base reaction (1 Viewer)

[Fiction]

Hello~
For the following equation
H2So4 + 2NaOH --> 2H2O + NaSO4

With 0.086M and 35ml of h2so4 and 1.00m and 25ml of Naoh , how would you calculate moles of h remaining?

Not going to lie, it haven't got a clue in how to solve the question. At the moment I'm guessing would it be like calculate moles of h2so4 or NaOHband contrast it with h2o? Also what does it mean by h or h ions remaining? I thought all the hs reacted to form water since its a complete reaction

Thank you!

 

[Librah]

Hello~
For the following equation
H2So4 + 2NaOH --> 2H2O + NaSO4

With 0.086M and 35ml of h2so4 and 1.00m and 25ml of Naoh , how would you calculate moles of h remaining?

Not going to lie, it haven't got a clue in how to solve the question. At the moment I'm guessing would it be like calculate moles of h2so4 or NaOHband contrast it with h2o? Also what does it mean by h or h ions remaining? I thought all the hs reacted to form water since its a complete reaction

Thank you!

Acids must undergo ionisation (release H+) to actually contribute to acidity of a solution since pH=-log[H+]. Same as Bases in releasing OH-.
I'll assume your not up to Bronsted Lowry yet. Well in reality, here if H2SO4 (Diprotic-Has 2 hydrogens that can be released as ions) is in excess, it well only completely undergo it's first ionisation H2SO4--->HS04+H+. Then partially ionise again releasing some more H+. Though i think in the HSC they just assume full ionisation for a H2SO4 since it's a strong acid.

Basically your trying to find the left over H+ when excess H2SO4 neutralises NaOH. So, assuming full ionisation H2SO4--->2H(+) + SO4(2-). (assume H2SO4 releases 2 H+ for every mole of it)
n(H2SO4):n(H+)=1:2
n(H+)=0.086x0.035x2
n(OH-)=1.00x0.025 <----- limiting
H+ +OH- --->H20 (neutralisation) Reacts in a 1:1 ratio.
So n(H+) remaining in excess = 0.086x0.035x2 - (1.00x0.025)

 

[Fiction]

Acids must undergo ionisation (release H+) to actually contribute to acidity of a solution since pH=-log[H+]. Same as Bases in releasing OH-.
I'll assume your not up to Bronsted Lowry yet. Well in reality, here if H2SO4 (Diprotic-Has 2 hydrogens that can be released as ions) is in excess, it well only completely undergo it's first ionisation H2SO4--->HS04+H+. Then partially ionise again releasing some more H+. Though i think in the HSC they just assume full ionisation for a H2SO4 since it's a strong acid.

Basically your trying to find the left over H+ when excess H2SO4 neutralises NaOH. So, assuming full ionisation H2SO4--->2H(+) + SO4(2-). (assume H2SO4 releases 2 H+ for every mole of it)
n(H2SO4):n(H+)=1:2
n(H+)=0.086x0.035x2
n(OH-)=1.00x0.025 <----- limiting
H+ +OH- --->H20 (neutralisation) Reacts in a 1:1 ratio.
So n(H+) remaining in excess = 0.086x0.035x2 - (1.00x0.025)

Thanks just a few quick questions, how did you get the n(OH) = 1.00*0.025? Also what would a base ionising equation look like - is it the same as acid but like replace h with OH?
Lastly, I don't fully understand how because h+OH ---> h2o reacts in a 1:1 ratio, we hence arrive at the conclusion. I was wondering, what happens if the H+OH wasn't in that 1:1 ratio? Like say if it was in a 2:1 ratio, would you times n(h+) by 2..?

Thanks again <3

 

[Librah]

Thanks just a few quick questions, how did you get the n(OH) = 1.00*0.025? Also what would a base ionising equation look like - is it the same as acid but like replace h with OH?
Lastly, I don't fully understand how because h+OH ---> h2o reacts in a 1:1 ratio, we hence arrive at the conclusion. I was wondering, what happens if the H+OH wasn't in that 1:1 ratio? Like say if it was in a 2:1 ratio, would you times n(h+) by 2..?

Thanks again <3

1. From your data. n(naoh)=n(oh-)= 1.00x0.025. NaOH-->Na+ +OH-
2. Thats just the general neutralisation reaction to complete a water molecule. It's like doing a jigsaw puzzle and fitting the parts. It just isn't a 2:1 cause nature says so/ the structure/bonding won't allow it, you could try to balance an equation like that, but it would be impossible 2H+ +OH- ---> ?????? . You times n(H+) in the example by 2 since H2SO4 is diprotic (has 2 hydrogens, and can thus release 2 hydrogen ions for every H2SO4 molecule, well we assume it can do that.)

 

[Fiction]

1. From your data. n(naoh)=n(oh-)= 1.00x0.025. NaOH-->Na+ +OH-
2. Thats just the general neutralisation reaction to complete a water molecule. It's like doing a jigsaw puzzle and fitting the parts. It just isn't a 2:1 cause nature says so/ the structure/bonding won't allow it, you could try to balance an equation like that, but it would be impossible 2H+ +OH- ---> ?????? . You times n(H+) in the example by 2 since H2SO4 is diprotic (has 2 hydrogens, and can thus release 2 hydrogen ions for every H2SO4 molecule, well we assume it can do that.)

@1: Thank you!
@2: I'm aware neutralisation won't go in the ratio 2:1 but I meant 'if' as in if this was another equation where this step has a 2:1 ration, then how would you work out the solution. This is because I don't fully understand how you integrated that step into your working and it's significance.
Thanks in advance to anyone who cares to answer

 

[Librah]

@1: Thank you!
@2: I'm aware neutralisation won't go in the ratio 2:1 but I meant 'if' as in if this was another equation where this step has a 2:1 ration, then how would you work out the solution. This is because I don't fully understand how you integrated that step into your working and it's significance.
Thanks in advance to anyone who cares to answer

Oh that was just telling you where some of the H+/OH- disappeared to, didn't have any relevance to the actual working out. Still a bit unsure what your asking.