Home
Class 12
CHEMISTRY
Al(OH)(3) darr "(white)" + OH^(-) to [Al...

`Al(OH)_(3) darr "(white)" + OH^(-) to [Al(OH)_(4)]^(-)` soluble complex
White precipitate of `Al(OH)_(3)` reappears when

A

a solution of ammonium chloride is added

B

a solution of ammonia is added

C

concentrated `HNO_(3)` is added in excess

D

all of these

Text Solution

Verified by Experts

The correct Answer is:
A

`underset("white ppt")(Al(OH)_(3))+OH^(-) to underset("soluble")([Al(OH)_(4)]^(-))`
`[Al(OH)_(4)]^(-)+NH_(4)^(+) to Al(OH)_(3) darr +H_(2)O+NH_(3) uarr`
Promotional Banner

Topper's Solved these Questions

  • PRACTICAL CHEMISTRY

    AAKASH SERIES|Exercise PRACTICE SHEET-1 (Integer answer type questions)|7 Videos
  • PRACTICAL CHEMISTRY

    AAKASH SERIES|Exercise PRACTICE SHEET-2 (Single answer questions)|15 Videos
  • PRACTICAL CHEMISTRY

    AAKASH SERIES|Exercise PRACTICE SHEET-1 (More than one correct answer Questions)|6 Videos
  • POLYMERS

    AAKASH SERIES|Exercise OBJECTIVE EXERCISE - 4|59 Videos
  • PRINCIPLES OF METALLURGY

    AAKASH SERIES|Exercise SUBJECTIVE EXERCISE-2 (LONG ANSWER QUESTIONS )|7 Videos

Similar Questions

Explore conceptually related problems

Al(OH)_(3) darr "(white)" + OH^(-) to [Al(OH)_(4)]^(-) soluble complex Fe(OH)_(3) precipitate and Al(OH)_(3) precipitate can be separated by

Al(OH)_(3) darr "(white)" + OH^(-) to [Al(OH)_(4)]^(-) soluble complex Identify the correct statement with respect to the gelatinous white precipitate of aluminum hydroxide.

CuSO_(4) +NH_(4)OH gives deep blue complex of

A colloidal sol Fe(OH)_(3) in water is

Al_(2)O_(3) to AIN to Al(OH)_(2) to Al_(2)O_(3) . The sequence of these products involved in

Al_(2)O_(3) to AlN to A(OH)_(3) to Al_(2)O_(3) . The sequence of these products involved in

The solution of a chemical compound reacts with AgNO_(3) solution to from a white precipitate of Y which dissolves in NH_(4)OH to give a complex Z. When Z is treated with dilute HNO_(3) , Y reappears. The chemical compound X can be:

Fe(OH)_(3) can be separated from Al(OH)_(3) by addition of:

AI(OH)_(3) is insoluble in excess of NH_(4)OH but soluble in excess of NaOH. Explain.

Why AI(HO)_(3) is insoluble in excess of NH_(4)OH but soluble in NaOH ?