What is #pH# if #[H_3O^+]=0.001molL^-1#?

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Answer 1 · 2017-06-08T11:48:38

#pH=3............#

#pH#, #"pouvoir hydrogene"#, #=-log_10[H_3O^+]#

Now it is a fact that in water the following autoprotolysis takes place:

#2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+#

The ionic product at #298*K# #=# #K_w=[H_3O^+][HO^-]=10^-14#.

If we take #-log_10# of both sides we get:

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

But by definition, #pH =-log_10[H_3O^+]# and #pOH =-log_10[HO^-]#

Thus #pK_w =14=pH+pOH#

But nitric acid should be stoichiometric in #H_3O^+# and #NO_3^-# ion...........

#HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-#

And so #pH# #=# #-log_10(10^-3)# #=# #3#, and #pOH=??#

And #pH=3#.

Capisce?

What is #pH# if #[H_3O^+]=0.001*mol*L^-1#?