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Question
Assuming complete dissociation, calculate the pH of the following solution:
0.005 M NaOH
Solution
\[\ce{NaOH_{(aq)} ↔ Na^+_{(aq)} + HO^-_{(aq)}}\]
`["HO"^-] = ["NaOH"]`
`=> ["HO"^-] = 0.005`
`"pOH" = - log["HO"^(-)] = -log(0.005)`
pOH = 2.30
∴ pH = 14 - 2.30
= 11.70
Hence, the pH of the solution is 11.70.
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