Acid radical analysis
A water soluble salt, when treated with diluted H 2 SO 4 , gives a colorless gas that turns lime water milky. On passing this gas in excess, the milkyness disappears. On treating the gas with acidic potassium dichromate, the color of potassium dichromate turns green. Find the molecular mass of the anionic part of the salt.
The answer is 80.
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2 solutions
Nice solution.
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I have edited my previous question. You can now check your answer
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I accidently put wrong answer and all my three chances have gone!
I would like to comment that the "it" in the question should have been more clear. The "it" could refer to the gas, or the salt. The English was extremely confusing. Please do take note and be clearer in the future :)
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I am sorry, it has been edited now. Thanks for your feedback.
It could also be Br- ion. All the above conditions are satisfied. It has atomic mass 80 too. That is the one I was considering when I entered the answer correctly.
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B r 2 doesn't turn Lime water milky.
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It releases SO2 gas when conc sulphuric acid is added since it can reduce SO4(2-) to SO2 gas. The rest is same as in your solution.
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@Rohan Rao – You dont understand, B r 2 doesn't turn lime water milky (it is one of the conditions), in fact, it disproportionates when it is passed through C a ( O H ) 2 .
B r 2 + C a ( O H ) 2 → C a B r 2 + C a ( B r O 3 ) 2 + H 2 O
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@Anish Puthuraya – What I meant is that along with Br2, there is a side product of SO2 that is formed, so it could turn the limewater milky.
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@Rohan Rao – That is correct, indeed. But since there are 2 gases liberated, it might create some confusion...though you are not wrong.
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@Anish Puthuraya – Yep, and there is definitely nothing contradicting SO3(2-) ion. Thanks for the reaction though, I didn't know that one would happen...never tried passing bromine through lime water...;)
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@Rohan Rao – That reaction is, in fact, true for any halogen and any strong base, except F 2 . The general reaction is :
For
Hot and Conc.
base,
X
2
+
O
H
−
→
X
−
+
X
O
3
−
+
H
2
O
and
For
Cold and Dil.
base,
X
2
+
O
H
−
→
X
−
+
O
X
−
+
H
2
O
Maybe, I will post a note on it someday.
@Rohan Rao – I am sorry if I am intruding but in the question t is mentioned that dilute sulphuric acid is added.
What about a hydrogen sulphate radical? It will still give sulphur dioxide.
From tests given in the question that the gas evolved turns lime water milky and it turns acidic potassium dichromate green, we can conclude that the gas evolved is sulphur dioxide i.e. S O 2 Now, we know that when a sulphite is treated with d i l . H 2 S O 4 sulphur dioxide is evolved. Therefore the anion in the given salt is S O 3 2 − As we know the atomic mass of sulphur is 3 2 and the atomic mass of oxygen is 1 6 Thus we need to find the molecular mass of this ion which is = 1 × 3 2 + 3 × 1 6 = 3 2 + 4 8 = 8 0
Only 2 gases turn lime water milky, C O 2 and S O 2
Now,
C O 2 is pretty stable in itself, as carbon has achieved its maximum valency, and would not undergo any oxidation. But, S O 2 on the other hand, is capable of oxidizing and will thus reduce Potassium Dichromate as following,
H + + S O 2 + C r 2 O 7 2 − → S O 4 2 − + C r 3 + (Green) + H 2 O
Thus,
Our required gas is S O 2
Now,
S O 3 2 − + H 2 S O 4 → H 2 O + S O 2 + S O 4 2 −
Hence, the anionic part of the water soluble salt is S O 3 2 − , whose molecular mass is 3 2 + 1 6 × 3 = 8 0 g