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