What is the #pH# of a solution of sulfuric acid of #5.98molL^-1# concentration?

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Answer 1 · 2016-06-22T04:24:18

#pH=-log_(10)[H_3O^+]#

#pH=-log_(10)[2xx5.98]# #~=-1#.

Why did I multiply the given concentration by 2?

What is the #pH# of a solution of sulfuric acid of #5.98*mol*L^-1# concentration?