Given that pure water has #pH=6.630# at #323*K#, what is #pK_w# at this temperature?

1 Answer
anor277
Nov 5, 2017

Good question.....we interrogate the equilibrium....

#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#

Explanation:

Now #K_w=[H_3O^+][HO^-]=10^(-14),# #298*K#....

#K_w=10^(-14),# #298*K#....

We want #K'_w,# #323*K#

But we got a handle on #[H_3O^+]#, i.e. #pH=6.630#....and since by definition, #pH=-log_10[H_3O^+]#, #H_3O^+=10^(-6.630)-=[HO^-]#

And thus #K'_w=10^(-6.630)xx10^(-6.630)=5.50xx10^-14#...and #pK_w'=13.26#...

Note that #K'_w>K_w# is reasonable, given that the autoprotolysis reaction as written is a bond-breaking process, which should be slightly favoured, i.e. towards the products as written, at elevated temperatures.....

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