Calculate pH of a buffer prepared by adding 10 mL of 0.10 M aceti

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

NH4Cl (s) is heated in a test tube. Vapours are brought in contact with red litmus paper, which changes it to blue and then to red. It is because of

  • formation of NH4OH and HCl

  • formation of NH3 and HCl

  • greater diffusion of NH3 than HCl

  • greater diffusion of HCl than NH3


2.

When 1 mole of CO2 (g) occupying volume 10 L at 27°C is expanded under adiabatic condition, temeprature falls to 150 K. Hence, final volume is

  • 5 L

  • 20 L

  • 40 L

  • 80 L


3.

Which of the following ligands is tetradentate?


4.

A near UV photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength 760 nm. Hence, wavelength of the second photon is

  • 1060 nm

  • 496 nm

  • 300 nm

  • 215 nm


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5.

A constant current of 30 A is passed through an aqueous solution of NaCl for a time of 1.00 h. What is the volume of Cl2 gas at STP produced?

  • 30.00 L

  • 25.08 L

  • 12.54 L

  • 1.12 L


6.

If for the cell reaction,

Zn + Cu2+  Cu + Zn2+;

entropy change S° is 96.5 J mol-1K-1, then temperature coefficient of the emf of a cell is

  • 5 × 10-4 VK-1

  • 1 × 10-3 VK-1

  • 2 × 10-3 VK-1

  • 9.65 × 10-4 VK-1


7.

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum?

  • n = 4 to n = 2

  • n = 3 to n = 2

  • n = 2 to n = 1

  • n = 4 to n = 3


8.

What is the degenercy of the level of H-atom that has energy -RH9 ?

  • 16

  • 9

  • 4

  • 1


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9.

What is the EAN of [Al(C4O4)3]3- ?

  • 28

  • 22

  • 16

  • 10


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10.

Calculate pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.1 M sodium acetate [ pKa (CH3COOH) = 4.74 ]

  • 3.00

  • 4.44

  • 4.74

  • 5.04


D.

5.04

[CH3COOH] = millimoles of CH3COOH

                  = 0.1 × 10 = 1.0

[CH3COONa]= millimoles of CH3COONa

                   = 0.1 × 20 = 2.0

From, Henderson Hasselbalch equation,

pH = pKa + log conjugate baseacid

    = 4.74 + log 21

    = 4.74 + 0.30 = 5.04


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