First let me say that this is one of the worst stated problems I've seen in years. It could be answered easily IF we knew which base was used. Since the base is not given we can look at two scenarios. The first is with a monobasic base and the second with a dibasic base. Take NaOH as the first one and Ba(OH)2 for the second.
1. H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
If you titrate 0.2 M H2SO4 with 0.2 M NaOH you see that it take twice as much NaOH to titrate the same amount of H2SO4 BECAUSE H2SO4 has TWO H ions to neutralize but NaOH has only ONE OH to do that so it takes twice as much NaOH to do the job. Therefore, it you start with 25 mL of H2SO4 you know it will take twice that much (50 mL) of the same molarity NaOH.
2. If we use Ba(OH)2 as the base then
H2SO4 + Ba(OH)2 ==> BaSO4 + 2H2O.
You see here that the ratio of acid to base is 1:1 so if you take 25 mL of H2SO4 you will titrate it with 25 mL Ba(OH)2.
Another note: Please note that you used the lower case m. Note that m stands for molality while M stands for molarity They aren't the same thing. You want to titrate with solution of known M and not m.
M stands for moles/liter of solution while m stands for moles/kg solvent.
Another note: We could have taken a tribasic base such as Al(OH)3 in which case the ratio is 3 acid:2 base as follows:
3H2SO4 + 2Al(OH)3 ==> Al2(SO4)3 + 6H2O so using 30 mL H2SO4 to start will require 30 x (2/3) = 20 mL of the base.
A solution of 0.2 m sulfiric acid is titrated with a 0.2 m basic solution. In three to five sentences, explain how you can use the results of the titration as evidence for the ratio of sulfiric acid and the base in the balanced chemical reaction for the titration.
y'all pls help i do not understand how this works at allll. i have literally gone back in my lessons and rewatched livelessons and everything tryna figure this out but i still just don't get it 😩
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