What is #pH# if #(i)# #[H_3O^+]=5.6xx10^-5molL^-1#; and #(ii)# if #[HO^-]=10^-7molL^-1#?

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Answer 1 · 2017-06-25T20:06:50

#(i)# #pH=4.25.................#

By definition, #pH=-log_10[H_3O^+]#, and thus for (i)....

#pH=-log_10(5.6xx10^-5)=-(-4.25)=4.25# (if I have counted up the zeroes properly........)

But since #[H_3O^+]xx[HO^-]=10^-14#, we can take #log_10# of both sides, and gets........

#log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14=-14#

Equivalently, multiplying each side by #-1#............

#underbrace(-log_10[H_3O^+])_color(red)(pH)underbrace(-log_10[HO^-])_color(blue)(pOH)=+14#

And thus our defining relationship......

#color(red)(pH)+color(blue)(pOH)= 14#

And so (ii) if #[HO^-]=10^-7*mol*L^-1#, #pOH=7# (why?), and #pH=7#. Agreed?

What is #pH# if #(i)# #[H_3O^+]=5.6xx10^-5*mol*L^-1#; and #(ii)# if #[HO^-]=10^-7*mol*L^-1#?