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प्रश्न
Assuming complete dissociation, calculate the pH of the following solution:
0.002 M KOH
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उत्तर
\[\ce{KOH_{(aq)} ↔ K^+_{(aq)} + OH^-_{(aq)}}\]
`["OH"^-] =["KOH"]`
`=> ["OH"^-] = .002`
Now `"pOH" = - log["OH"^-]`
= 2.69
`therefore pH = 14 - 2.69`
= 11.31
Hence, the pH of the solution is 11.31
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