Ph of 0.05m of h2so4..?
Answers
Answered by
2
hay!!
it's - H2SO4=2H+So4²-
so in 0.05m H2SO4
______________________________
Solution
Concentration of H+ =0.01m
Now phone of the solution will b -log(H+)
=-(-1)
=1
I hope it's help you
it's - H2SO4=2H+So4²-
so in 0.05m H2SO4
______________________________
Solution
Concentration of H+ =0.01m
Now phone of the solution will b -log(H+)
=-(-1)
=1
I hope it's help you
Anonymous:
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Is there any other way to do it
Answered by
0
Concentreation of H = + 0.01m
.........
pH of solution will be log ( H + )
=> - ( - 1 )
=> 1
...........
so in 0.05m is H2SO4 ( sulphuric acid )
.........
pH of solution will be log ( H + )
=> - ( - 1 )
=> 1
...........
so in 0.05m is H2SO4 ( sulphuric acid )
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