What is the pH of #1 * 10^-4# #M# #NaOH#?

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Answer 1 · 2016-05-28T07:03:44

#pH = 10#

In water the following equilibrium operates:

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

As for any equilibrium, we can write the equilibrium expression:

#([H_3O^+][HO^-])/[[H_2O]]# #=# #K'_w#

Since #[H_2O]# is huge, we can simplify this expression:

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

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

Taking #log_10# of both sides:

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

Now we can define #pH# #=# #-log_10[H_3O^+]#, and

#pOH# #=# #-log_10[HO^-]#.

And thus #pOH +pH=14# #"(finally!)"#

Thus #pOH# of #1xx10^-4mol*L^-1" NaOH"# #=# #-log_10(1xx10^-4)=4#

Since #pOH +pH=14#, #pH=10#, as required.