When an aqueous solution at room temperature is analyzed, the #[H^+]# is found to be #2.0*10^-3# #M#. What is the #[OH^-]#?

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Answer 1 · 2016-08-13T16:55:18

#[HO^-]# #=# #5.01xx10^-12*mol*L^-1#

As you know water undergoes autoprotolysis:

#2H_2O rightleftharpoons H_3O^+ + HO^-#

At #298*K# this equilibrium has been measured exhaustively and precisely:

#K_w=[H_3O^+][HO^-]# #=# #10^-14#

If we take #-log_10# of both sides:

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

i.e. #14=pH+pOH#, because #pH=-log_10[H_3O^+]# and #pOH= -log_10[HO^-]# by definition.

So #[H_3O^+]# #=# #2.0xx10^-3#, and #pH=-log_10(2.0xx10^-3)# #=# #-(-2.70)=2.70#.

Thus #pOH=14-2.70=11.30#. #[HO^-]# #=# #10^(-11.30)*mol*L^-1=5.01xx10^-12*mol*L^-1#.

You can check to see if I am right by multiplying #[H_3O^+]# and #[HO^-]#. What should the product equal?