A weak acid of dissociation constant 10-5 is being titrated with

Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

In Periodic Table, the basic character of oxides

  • increases from left to right and decreases from top to bottom

  • decreases from right to left and increases from top to bottom

  • decreases from left to right and increases from top to bottom

  • decreases from left to right and increases from bottom to top


12.

Which one of the following contains P-O-P bond?

  • Hypophosphorus acid

  • Phosphorus acid

  • Pyrophosphoric acid

  • Orthophosphoric acid


13.

Which of the following orbitals will have zero probability  of finding the electron in the yplane. 

  • px

  • py

  • pz

  • dyz


14.

What type of orbital hybridisation is considered on P in PCl5?

  • sp3d

  • dsp3

  • sp3d2

  • d2sp3


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15.

In haemoglobin, the metal ion present is

  • Fe2+

  • Zn2+

  • Co2+

  • Cu2+


16.

The bond order of CO molecule is

  • 2

  • 2.5

  • 3

  • 3.5


17.

Which of the following orders regarding ionisation energy is correct?

  • N > O > F

  • N < O < F

  • N > O < F

  • N < O < F


18.

Which of the following compounds shows evidence of the strongest hydrogen bonding?

  • Propan-1-ol

  • Propan-2-ol

  • Propan-1,2-diol

  • Propan-1,2,3-triol


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19.

Identify [A] and [B] in the following

Ac89227 -β [A] -α [B] -α Rn

  • Po, Rn

  • Th, Po

  • Ra, Th

  • Th, Ra


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20.

A weak acid of dissociation constant 10-5 is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be

  • 5 + log 2 - log 3

  • 5 - log 2

  • 5 - log 3

  • 5 - log 6


B.

5 - log 2

The pH at the point of one-third neutralisation of the acid will be calculated as

Ka = 10-5 ; pKa = -log Ka = -log 10-5 = 5

Initial Final  HA1 mole1 - 13mol= 23 mol + NaOH0  NaA013 mol13 mol + H2O013 mol

 

Here, final solution acts as an acidic buffer.

pH = pKa + log saltacid

or, pH = 5 + log 1323

          = 5 + log 12

 pH  = 5 - log 2


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