What is the #pH# of an aqueous solution prepared from a #600mg# mass of #NaOH# dissolved in #75mL# of water?

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Answer 1 · 2016-10-27T08:38:39

#pH~=13#

We need to find #[HO^-]# #=# #"Moles of hydroxide"/"Volume of water"# #=# #(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)# #=# #0.20*mol*L^-1#.

Now #pOH# #=# #-log_10[HO^-]# #=# #-log_10(0.20)# #=# #-(-0.70)# #=# #0.70#.

But we know (or should know) that in water at #298*K#, #pH+pOH=14#.

Thus #pH=14-0.70=?????????#

What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water?