hey friends what is PH of 10^-8N h2so4?
it's not 8
plzzzz help me!!! ❤
Answers
Answered by
1
The 'n' factor of H2SO4 is 2 and
Normality= Molarity x n factor
Therefore,
Molarity=10^-8/2
=5x10^-9M
As pH=-log[H+]
Therefore, pH=-log[5x10^-9]
So, pH=9.698
But the pH of H2SO4 can't be greater than 7 as it is an acid.
So, we have to take into account the ionisation constant of water too..
Therefore,
Total H+ions
= 10^-7+5x10^-9
=1.005x10^-7
Therefore pH=-log[1.005x10^-7]
≈7-0.2
≈6.8
Normality= Molarity x n factor
Therefore,
Molarity=10^-8/2
=5x10^-9M
As pH=-log[H+]
Therefore, pH=-log[5x10^-9]
So, pH=9.698
But the pH of H2SO4 can't be greater than 7 as it is an acid.
So, we have to take into account the ionisation constant of water too..
Therefore,
Total H+ions
= 10^-7+5x10^-9
=1.005x10^-7
Therefore pH=-log[1.005x10^-7]
≈7-0.2
≈6.8
abhinav27122001:
In which area
ND ur
kyA hua
Ohk
I live in Mumbai
is the chapter of Ionic Equilibrium completed in your classes
:-))
yess
see you know that for the pH to B7 the concentration of h + Ion should be 10 raise to minus 7
and the value of concentration of h + ions which you have mentioned is less than 10 raise to minus 7 so the pH has to close to 7 as it is an acid it can't be greater than 7
Answered by
0
PH=6.97 It is correct.
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