2 H₂O (l) ⇒ H3O*(aq) + OH(aq) Kw = 2.92 x 10-14 at 40 °C. Calculate the pH of 0.10 M NaOH at this temperature. 13.00 14.54 12.04 12.53 O 14.04

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2 H₂O (1) H3O+ (aq) + OH¯(aq) Kw=
Calculate the pH of 0.10 M NaOH at this temperature.
13.00
14.54
12.04
12.53
= 2.92 x 10-14 at 40 °C.
14.04
Transcribed Image Text:2 H₂O (1) H3O+ (aq) + OH¯(aq) Kw= Calculate the pH of 0.10 M NaOH at this temperature. 13.00 14.54 12.04 12.53 = 2.92 x 10-14 at 40 °C. 14.04
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Why do we use the kw from this equation 2H2O(l) ⇄⇄H3O+(aq) + OH-(aq) to find the OH- of NaOH, instead of NaOH⇄⇄Na+ + OH- ?

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