SODIUM HYDROXIDE (NaOH) MCQ
IIT JEE - NEET

AdiChemistry IIT JEE

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1) Aqueous sodium hydroxide reacts with white phosphorous to form phosphine and :

(EAMCET 2001 M)

1) NaH2PO2 

2) P2O5 

3) Na2PO3 

4) P2O3 

Logic and Solution:

When phosphorus dissolves in caustic soda (NaOH), phosphine (PH3) gas is liberated along with formation of sodium hypophosphite (NaH2PO2).

3NaOH + P4 + 3H2O ---------> 3NaH2PO2 + PH3

Conclusion:

The correct option is '1'.

Related questions

2) Which of the following pairs liberate gas a when they react with each other?

(EAMCET 2005 M) 

1) SiO2, Na 

2) Fused NaOH, C 

3) Mg, B2O3 

4) Mg, CO2

Logic & solution: 

1) There is no reaction between silica, SiO2 and sodium metal, Na.

2) Fused NaOH reacts with carbon to liberate dihydrogen gas.

6NaOH + 2C ----------> 2Na2CO3 + 3H2 

3) Magnesium reduces boric anhydride to give boron.

3Mg + B2O3 ----------> 3MgO + 2B

4) Magnesium reduces carbon dioxide to carbon.

2Mg + CO2 ---------> 2MgO + C

Conclusion:

The correct option is: 2. 

3) Zn2+ dissolves in excess of NaOH due to the formation of:

(EAMCET 1992 M) 

1) Zn(OH)2 

2) ZnO 

3) Na2ZnO2 

4) all the above

Logic & solution: 

In excess of sodium hydroxide, Zn2+ forms sodium zincate, Na2ZnO2 that is soluble in water. It is a complex salt and should be represented by the formula: Na2[Zn(OH)4].

Zn2+ + 2NaOH (excess) ----------> Na2ZnO2 + 2H+

or

Zn2+ + 4NaOH (excess) ----------> Na2[Zn(OH)4] + 2Na+

However, when small amount of sodium hydroxide is added, a while gelatinous precipitate of Zn(OH)2 is formed.

Zn2+ + 2NaOH (little) ----------> Zn(OH)2 + 2Na+

Conclusion:

The correct option is: 3. 

4) The Castner-Kellner cell used for the manufacture of NaOH, the cathode in the central compartment is made of:

(EAMCET 1992 M) 

1) Iron 

2) Carbon 

3) Mercury 

4) Nickel 

Logic & solution: 

In the Castner-Kellner cell, used to manufacture sodium hydroxide, a bunch of iron rods is used as cathode in the central compartment.

In the side compartments, graphite rods are used as anodes.

Mercury acts as cathode in side compartments and as anode in the middle compartment.

Conclusion:

The correct option is: '1'. 

5) Consider the reaction: 6NaOH +3Cl2 -------> 5NaCl + A + 3H2O. The oxidation number of chlorine in “A” is:

(EAMCET 2001 M) 

1) Iron 

2) Carbon 

3) Mercury 

4) Nickel 

Logic & solution: 

'A' is: NaClO3. The oxidation number of Cl is +5. 

Calculation:

Na + Cl + 3O = 0

(+1) + Cl + 3(-2) = 0

Cl = +5

Conclusion:

The correct option is: '1'. 

 

 

AdiChemistry IIT JEE

(new) Click here to see 3d Interactive Solved Question paper 

Author: Aditya vardhan Vutturi