Two 20.0-mL samples, one 0.200 M KOH and the other
0.200 M CH3NH2...

Question

Two 20.0-mL samples, one 0.200 M KOH and the other
0.200 M CH3NH2, are titrated with 0.100 M HI.
a. What is the volume of added acid at the equivalence point
for each titration?

Accepted Answer

Hello. Everyone in this video we're trying to calculate for the volume asset that should be added to reach the equivalence point for each dehydration. And were taken into consideration the basic solutions of N A O H And C six H 5 and H two. So let's go ahead and recall that at the equivalence point arbitration, that the most of acid well equal to our moles of base. And of course we should know that the polarity is equal to the most over volume isolate our moles here because that's our interest is most is equal to the mill, arat e times volume. All right now putting this information with this together. Then we should know that we mean our malaria times volume of our acid equal to the malaria times volume of our base. All right. So the volume of acid is equal to them. The military times volume of our base over the malaria t of our acid. Putting numerical values into this equation, then We'll have 25.0 ml Times are 0.450 moller divided by the milan city of our acid, which is given to us as 0. smaller, screw them down a little. Then the volume of our assets Is equal to 75.0 mm. And that's actually going to be for both situations. All right. So then The volume of asset that needs to be added to reach the equivalent point for both tight rations is going to equal to 75.0 ml. That's going to be my final answer for this problem. Thank you all so much for watching.

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH3NH2, are titrated with 0.100 M HI. a. What is the volume of added acid at the equivalence point for each titration?

Watch next