How do we differentiate between `Fe^(3+)` and `Cr^(3+)` ions in group III ?A. By adding excess of `NH_(4)OH` solutionB. By increasing `NH_(4)^(+)` ion concentrationC. By decreasing `OH^(-)` ion concentrationD. Both (b) and (c) are correct

How do we differentiate between `Fe^(3+)` and `Cr^(3+)` ions in group

1.	How do we differentiate between `Fe^(3+)` and `Cr^(3+)` ions in group III ?A. By adding excess of `NH_(4)OH` solutionB. By increasing `NH_(4)^(+)` ion concentrationC. By decreasing `OH^(-)` ion concentrationD. Both (b) and (c) are correct
Answer» Correct Answer - B `k_(sp)` value of `Fe(OH)_(3)` is less that of `Cr(OH)_(3)`. Therefore, the concentration of `OH^(-)` ions in the solution, can be limited by either increasing `NH_(4)^(+)` ion concentration or by decreasing `OH^(-)` ion concentration

Discussion

No Comment Found

Related InterviewSolutions

A metal nitrate reacts with KI solution to give a block precipitate which on addition of excess of KI solution forms an orange coloured solution. The cation of metal nitrate is :A. `Hg^(2+)`B. `Bi^(3+)`C. `Pb^(2+)`D. `Cu^(2+)`
A solution of a metal ion when treated with KI solution gives a red precipitate which dissolves in excess of KI solution to give a colourless solution. Moreover, the solution of the same metal ion on treatment with the solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion isA. `Pb^(2+)`B. `Hg^(2+)`C. `Cu^(2+)`D. `Co^(2+)`
An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution, a black precipitate is obtained. The substance is :A. `Hg^(2+)` saltB. `Cr^(3+)` saltC. `Ag^(+)` saltD. `Pb^(2+)` salt
A white crystalline salt [A] reacts with dilute HCl to liberate a suffocatig gas [B] and also forms a yellow precipitate. The gas [B] turns potassium dichromatic acidified with dilute `H_(2)SO_(4)` to a green coloured solution [C]. The compound A, B and C are respectivelyA. `Na_(2)SO_(3), SO_(2), Cr_(2)(SO_(4))_(3)`B. `Na_(2)S_(2)O_(3), SO_(2), Cr_(2)(SO_(4))_(3)`C. `Na_(2)S, SO_(2), Cr_(2)(SO_(4))_(3)`D. `Na_(2)SO_(4), SO_(2), Cr_(2)(SO_(4))_(3)`
A compound is soluble in water. If ammonia is added to aqueous solution of compound, a reddish brown precipitate appears which is soluble in dill. HCl. The compound is a salt of :A. aluminiumB. zincC. ironD. cadmium
In the second group of qualitative analysis, `H_(2)S` is pass through a solution acidified with HCl in order to :A. limit the concentration of `S^(2-)` ionsB. increase the solubility of `H_(2)S`C. increase the concentration of `S^(2-)` ionsD. provide extra `Cl^(-)` ions
Dolute nitric acid is generally not used for the preparation of original solution for the basic radicals because it :A. it is a reducing agentB. it is an oxidising agentC. forms insoluble nitratesD. forms soluble nitrates
Silver, mercury (our) and lead are grouped toether in the same group of qualitative analysis because they form:A. nitratesB. carbonates which dissolve in dill. `HNO_(3)`C. insoluble chloridesD. colourless compounds
Identify the correct order of solubility of `Na_(2)S, CuS` and ZnS in aqueous solution :A. `CuS gt ZnS gt Na_(2)S`B. `ZnS gt Na_(2)S gt CuS`C. `Na_(2)S gt CuS gt ZnS`D. `Na_(2)S gt ZnS gt CuS`
A solution which is `10^(-3) M` each in `Mn^(2+), Fe^(2+), Zn^(2+)` and `Hg^(2+)` is treated with `10^(-16) M` sulphide ion. If `K_(sp)` of MnS, FeS, ZnS and HgS are `10^(-15), 10^(-23), 10^(-20)` and `10^(-54)` respectively, which one will precipitate first ?A. FeSB. MgSC. HgSD. ZnS

How do we differentiate between `Fe^(3+)` and `Cr^(3+)` ions in group III ?A. By adding excess of `NH_(4)OH` solutionB. By increasing `NH_(4)^(+)` ion concentrationC. By decreasing `OH^(-)` ion concentrationD. Both (b) and (c) are correct

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment