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प्रश्न
Assuming complete dissociation, calculate the pH of the following solution:
0.002 M HBr
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उत्तर
\[\ce{HBr + H_2O ↔ H_3O+ + Br-}\]
`["H"_3"O"^+] = ["HBr"]`
`=> ["H"_3"O"^+] = 0.002`
`therefore "pH" = - log ["H"_3"O"^+]`
= - log (0.002)
= 2.69
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