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Assuming complete dissociation, calculate the pH of the following solution: 0.003 M HCl

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प्रश्न

Assuming complete dissociation, calculate the pH of the following solution:

0.003 M HCl

संख्यात्मक
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उत्तर

\[\ce{H_2O + HCl ↔  H_3O+ + Cl-}\] 

Since HCl is completely ionized,

`["H"_3"O"^+] = ["HCl"]`

`=>["H"_3"O"^+] = 0.003`

Now

`"pH" = - log["H"_3"O"^+] = - log(0.003)`

= 2.52

Hence, the pH of the solution is 2.52.

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पाठ 6: Equilibrium - EXERCISES [पृष्ठ २३७]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
पाठ 6 Equilibrium
EXERCISES | Q 7.48 - (a) | पृष्ठ २३७
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