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EmilyParkinson
#1 Posted : Monday, July 16, 2012 5:41:32 PM
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could someone please explain why pH decreases as Kw increases with increasing temperature??
Kw=[H][OH] so if Kw increases, the product of [H][OH] would also have to increase.
pH = -log[H] -> increase [H] = more acidic
pOH = -log[OH] -> increase [OH] = more basic
if the pOH also decreases, 14-(small)pOH = (large)pH = basic ???

thank you!
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danielvis
#2 Posted : Monday, July 16, 2012 7:40:48 PM
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Hi Emily,

Everything you said during that explanation was correct, besides the assumption you made that as temperature increases, the sum of pH + pOH will still equal 14.

The sum of pH + pOH will be whatever the -log(Kw) is and therefore will change will the changing temperature.

So you are right that the pH and pOH both decrease, and both decrease by the same amount, but they will sum to a new value now.
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Kerolos_4875
#3 Posted : Tuesday, June 02, 2020 9:04:30 AM
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Hi! Looking at EK ICE 7 Q 158, I do not understand why we are using log(1/10) to represent the ratio of acetic acid to its conjugate base in the Henderson-Hasselbalch equation. I get that the question states we have a ratio of 10:1 (acetic acid to base) but why is it being represented this way? Thank you!
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INSTR_Katerina_102
#4 Posted : Sunday, June 07, 2020 6:45:19 PM
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Hi,

Recall that the Henderson Hasselbalch equation is pH = pKa + log([A-]/[HA]).

Because the acetic acid is HA, and the acetate ion conjugate base is A-, if we have 10 HA and 1 A- this equation becomes pH = pKa + log(1/10).
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