what will be the ph of 0.1 M solution of caustic potash?

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what will be the ph of 0.1 M solution of caustic potash?

2007-02-08 04:19:59 · 4 answers · asked by Rajeeva R 1 in Science & Mathematics ➔ Chemistry

4 answers

Huh, Steve_Geo gave a good and detailed calcualtion, but made a mistake:

Kw = [H+] * [OH-] = 10 ^-14

[OH-] = 0.1 = 10 ^-1

[H+] = Kw / [OH-] = 10 ^-14 / 10 ^-1 = 10 ^-13

Therefore,

pH = -log [H+] = 13, as Gervald F calculated.

2007-02-08 06:12:22 · answer #1 · answered by vjstrugar 2 · 0⤊ 0⤋

11-12

2007-02-08 04:23:09 · answer #2 · answered by Dirty Sanchez 3 · 0⤊ 1⤋

[KOH] = [OH-] = 0.1

Kw = [H+][OH-] = 1 x 10^-14

[H+](0.1) = 1 x 10^-14 = 10 x 10^-15

[H+] = 10^-14

pH = -Log[H+] = -Log 10^-14 = 14

2007-02-08 04:26:54 · answer #3 · answered by steve_geo1 7 · 0⤊ 0⤋

It will be 13.0 exactly, since it is a strong alkali.

2007-02-08 04:24:01 · answer #4 · answered by Gervald F 7 · 0⤊ 0⤋