Question

Important Question. Please explain with steps.

Answer 1

@Mani_Jha , @saifoo.khan , @heena , @supercrazy92 please help.

Answer 2

No chem! :(

Answer 3

Try at least.

Answer 4

@Mani_Jha is a boss at chem.

Answer 5

Yeah, that's why i called Mani Bhai also. He is my teacher. I hope he will do it this time too.:)))))

Answer 6

Let the concentration of NH3 be x.
Now, the resultant solution was basic(as it had to be neutralized by NaOH). During the reaction of neutralization:
M1V1=M2V2
x20=30*0.1
Find z.
Let the initial volume of NH3 be V.
Then when NH3 reacted initially with H2S04, 100 ml of H2S04 neutralized (v-20) ml of NH3(Because 20 ml was un-neutralized and made the solution basic)
Again use the law of equivalents in this case(be careful: the n-factor of H2S04 is 2)
Did you understand?
@Saifoo.khan, you just guessed that, didnt u?

Answer 7

Mani Bhai, can u show the second equation what we need to take?

Answer 8

(v-20)x=100*0.1*2. ok?

Answer 9

Okay, mani bhaiiiiiii

Answer 10

Wordings of these problems are terrible; teachers should bear in mind that students are not reading their minds and that wordings should be unambiguous.

In first problem, I find 17 millimole NH3 or 0.38 litre in STP. That holds if "equivalence" in the wording is meant for 'first equivalence', because eventually, NH4+ formed will also be titrated.

In second problem there should be 1.5 millimole of carbonate and 0.5 millimole of hydrogencarbonate.
Here again, problem should read: "The second end point was reached when AN EXTRA 20 ml of the acid was added

Answer 11

Oooookkkkkk, I will consult with my teacher about this. But these r the exact wordings that our teacher gave.

Answer 12

What about the answers?

Answer 13

btw, do you use 22.4 litres/mole or 22.7 litres/mole for molar volume in STP?

Answer 14

22.4

Answer 15

Do you have expected answers ?

Answer 16

Not now. i need to submit it as my assignment. i couldn't do it, so i asked it here.

Answer 17

Do my answers coincide with yours of Mani-Jha's?
Do you understand what reactions are at stake here?

Answer 18

My teacher gave the answer just now as 112 ml for the first one, and for the second one:
Weight of NaOH = 0.8g, weight of Na2CO3 = 1.59 g

Answer 19

1st one is correct. I forgot to divide something by 5, as you take only a 20 mL sample of the 100 mL solution.

So there were 5 millimoles of ammonia ie a volume of 112 mL

Answer 20

Thanks a lot

Answer 21

I have 159 mg of Na2CO3 which is not 1.59 g, unless concentration af acid is 1M instead of 0.1M.

Then I do not understand the NaOH part. Your solution is supposed to have NaHCO3.

What do you think?

Answer 22

Even I think the same.