Question

the PH value of 0.002M H2SO4 is

Answer 1

pH =-log(H+)
= -log(0.002)
=+3-0.3010
=2.699

Answer 2

The pH value of 0.002M $H_2SO_4$ is 2.69.

Given:

Molarity of the solution = 0.002M

To find:

pH of the solution

Formula to be used:

pH = -log[$H^+$]

Calculation:

When the given compound $H_2SO_4$ is completely dissociated it leads to the formation of 2 $H^+$ ions as per the following chemical equation:

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

As dissociation yields 2 $H^+$ ions, therefore

pH =  -log[$H^+$]

pH = -log[2]

pH = 2.69

Conclusion:

The pH value of 0.002M $H_2SO_4$ is 2.69.

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