calculate the (H3O+) &(HO-) at 298K in
(a) 0.01 M Hcl
(b) 0.005 MNAOH
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Answer:
(b) 0.005 M NaOH
NaOH(aq) ↔ Na+(aq) + HO-(aq)
[NaOH] = [ HO-]
⇒ [ HO-] = 0.05
pOH = -log[ HO-] = -log (0.05)
= 2.30
∴ pH = 14 - 2.30 = 11.70
Hence, the pH of the solution is 11.70
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