calculate and show the pH value of 0.02 M NaOH solution (log2 = 0.3010 , log5 = -0.6990).
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Answered by
5
Answer
NaOH(aq) ↔ Na+(aq) + HO-(aq)
[NaOH] = [ HO-]
⇒ [ HO-] = 0.02
pOH = -log[ HO-] = -log (0.02)
=-(log2×10^-2)
= 2-0.30
=1.7
∴ pH = 14 - 1.70 = 12.3
Hence, the pH of the solution is 12.3
Hope it helps u
Answered by
0
Answer: pH= 12.3 M
Explanation:
NaOH(aq) ↔ Na+(aq) + HO-(aq)
[NaOH] = [ HO-]
⇒ [ HO-] = 0.02
pOH = -log[ HO-] = -log (0.02)
=-(log2×10^-2)
= 2-0.30
=1.7
∴ pH = 14 - 1.70 = 12.3
Hence, the pH of the solution is 12.3
Hope it helps u
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