What is #pH# of a #0.002molL^-1# solution of #HCl(aq)#?

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Answer 1 · 2016-10-06T19:53:25

#pH=-log_10[H_3O^+]#

#[H_3O^+]# #=# #(10^-2*Lxx0.1*mol*L^-1)/(500*mLxx10^-3L*mL^-1)#

#=# #2.0xx10^-3mol*L^-1#

And #-log_10(2.0xx10^-3)# #=# #-(-2.70)=2.70#

What is #pH# of a #0.002*mol*L^-1# solution of #HCl(aq)#?