Page 81 - Chemistry - XI

obtained. To this solution, add acetic acid and lead acetate solution. A yellow precipitate is obtained.
This confi rms the presence of Cl ion (Chromyl chloride test).
–
4NaCl+ KCrO +3HSO → K SO +2CrOCl+ 2NaSO+3H O
2 2 7 2 4 2 4 2 2 2 4 2
CrOCl+ 4NaOH → NaCrO +2NaCl+2HO
2 2 2 4 2
Pb COOCH + NaCrO PbCrO +2CH COONa

3 2 4 4 3
Br and I do not give this test because Chromyl bromide and Chromyl iodide are not stable.
–
–
Q42. Why is fi ltrate of group IV concentrated before testing group V?
Ans. Group V cation can be precipitated only if concentration of group V cation is higher, therefore, it
is concentrated. Secondly, in dilute solution (NH ) CO gets hydrolysed forming H CO which is a
4 2 3 2 3
weak electrolyte and does not give suffi cient CO for precipitation.
2−
3
2−
Q43. Can we use sodium carbonate extract for detecting CO radical?
3
Ans. No, this is because sodium carbonate itself contains CO radical.
2−
3
Q44. Why should soda extract for S , SO , SO be acidifi ed before doing their confi rmatory test?
2−
2−
2–
4
3
Ans. It is used to dissolve carbonates because carbonates of Ag and Ba are insoluble which are used in
2+
+
their detection.
Q45. When carbonates of Ca , Ba , Sr , Pb are treated with dil. H SO , eff ervescence due to
2+
2+
2+
2+
4
2
evolution of Co is not brisk. Why?
2
Ans. Sulphates of Ca , Ba , Sr , Pb are heavy and deposited on carbonates and slow down the reaction.
2+
2+
2+
2+
That is why dil. H SO is used for detecting CO ion.
2−
4
2
3
Q46. Why does acidifi ed K Cr O paper turns green in the test of a sulphite ion?
2 2 7
Ans. Potassium dichromate is reduced to green coloured chromium sulphate by SO .
2

KCrO +3SO +H SO KSO+Cr SO +H O
2 2 7 2 2 4 2 4 2 4 3 2
green

Q47. What happens when BaCl is heated with Na CO in presence of water?
2 2 3
Ans. Barium chloride reacts with Na CO to form insoluble BaCO and NaCl formed is soluble in water.
2
3
3
BaCl +NaCO → BaCO+ 2NaCl
2 2 3 3
Q48. Why is yellow colloidal solution obtained on passing H S gas through group I fi ltrate if group
2
II is absent?
Ans. It is colloidal sulphur which is formed by oxidation of H S by oxidising agent.
2
HS+O HO +S
2 2
(Colloidal)
Q49. Can we use NaOH in place of NH OH in group III analysis? If not, why?
4
Ans. NaOH is strong base, therefore, higher group radicals will also get precipitated as their hydroxides,
therefore, it cannot be used.
Q50. How will you distinguish between ferrous and ferric salt?
Ans. Fe ions give green precipitate with NH OH whereas Fe gives reddish brown ppt. with NH OH.
3+
+2
4
4
Secondly, Fe decolourises acidifi ed solution of KMnO whereas Fe does not.
3+
+2
4
Q51. What is a reagent?
Reagent is a chemical substance which is used in qualitative and quantitative analysis.
Ans. Reagent is a chemical substance which is used in qualitative and quantitative analysis.
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