If the pH of a solution is 10.7, what would the pOH be?

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Answer 1 · 2016-10-06T16:04:26

3.3

Remember a simple thing that pH+pOH=14
so pOH= 14-10.7 = 3.3

Answer 2 · 2016-10-06T16:12:37

#pOH=3.3#

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

At #298*K#, we can write and quantify this equilibrium reaction:

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

And taking #log_10# of both sides:

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

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

But, by definitions, #pH=-log_(10)[H_3O^+]#, and #pOH=-log_(10)[HO^-]#, and #-logK_w=pK_w#.

And thus, #pK_w=14=pH+pOH#.

Given your question (finally we've got to it!), #pOH=14-pH# #=# #14-10.7=?#.