The pH of a 10-8 molar solution of HCl in water is from Chemistr

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 Multiple Choice QuestionsMultiple Choice Questions

81.

Equivalent amounts of H2 and I2, are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc at this temperature is :

  • 64

  • 16

  • 0.25

  • 4


82.

For the reaction H2(g) + I2(g)  2HI(g), the equilibrium constant KP changes with: 

  • total pressure

  • catalyst

  • the amount H2 and I2

  • temperature


83.

In a closed container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true ?

  • E = W 0, q= 0

  • E = W =q 0

  • E =0, W = q0

  • W =0, E= q0


84.

If an allotropic form changes slowly to a stable form. It is called

  • enantiotropy

  • dynamic allotropy

  • monotropy

  • None of the above


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85.

The pH of a buffer solution containing equal molal concentration of a weak base and its chloride (Kb for weak base = 2 x 10-5) is 

  • 5

  • 9

  • 4.7

  • 9.3


86.

Consider the following equilibrium in a closed container

        N2O4(g)  2NO2(g)

At a fixed temperature, the volume of the reaction container is halved. For this change which of the following statement holds true regarding the equilibrium constant (Kp) and degree of dissociation (α)?

  • Neither KP nor α changes

  • Both Kp and α changes

  • Kp changes but α does not

  • Kp does not change but α changes


87.

Conjugate acid S2O82-

  • H2S2O8

  • H2SO4

  • HS2O8-

  • HSO4-


88.

The equilibrium constant of a reaction is 300. If the volume of reaction flask is tripled, the equilibrium constant is

  • 300

  • 600

  • 900

  • 100


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89.

A 0.01 M ammonia solution is 5% ionised, its pH will be

  • 11.80

  • 10.69

  • 7.22

  • 12.24


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90.

The pH of a 10-8 molar solution of HCl in water is

  • 8

  • between 7 and 8

  • between 6 and 7

  • None of these 


C.

between 6 and 7

pH= -log[H+] = -log 10-8= 8

It is not possible for acid, so it is [H+], the [H+] of water is also added.
Total [H+] in solution

           =[H+] of HCl + [H+] of water= ( 1 × 10-8 + 1 × 10-7) M= ( 1+ 10) × 10-8 = 11 × 10-8M pH  = - log [H+]            = -log 11 × 10-8            =-log 11 + 8 log 10            = -1.0414 + 8           = 6.9586          

 


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