Ionic product of water at #310# #K# is #2.7# x #10^-14#. What is the #pH# of neutral water at this temperature?

1 Answer
anor277
Jan 17, 2016

Because acid base dissociation is a bond breaking reaction, and you have increased the temperature (thus favoring bond-breaking), #pH# should be less than #7#.

Explanation:

#H_2OrarrH^(+) + OH^-#; #K_w# #=# #[H^+][OH^-]# #=# #2.7xx10^(-14)# at #310# #K# (this is almost 3 times the value at #298# #K#).

By definition #pH=-log_10[H^+] # #=# #-log_(10){sqrt{2.7xx10^(-14)}}#. #pH# should be less than #7#, because the higher temperature facilitates the breaking of the #H-OH# bond.

Related questions
See all questions in pH calculations
Impact of this question
5519 views around the world
You can reuse this answer
Creative Commons License