Question:
The pOH of 0.0005 M sulphuric acid is
The pOH of 0.0005 M sulphuric acid is
Updated On: Jun 21, 2022
- 5
- 3
- 11
- 12
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The Correct Option is C
Solution and Explanation
$ { H_2SO_4 <=> 2H^+ +SO_4^{2-}}$
So, $ {[H^+] = 2\times 0.0005 = 1 \times 10^{-3} \, M }$
$ {pH = - log [H^+] = - log(1\times 10^{-3}) }$
$ {pH = 3 }$
$\because {pH +pOH = 14 }$
???$ {3 +pOH = 14 }$
????????$ { pOH = 14 - 3 }$
????????$ {pOH = 14 }$
????????$ {pOH = 11 }$
So, $ {[H^+] = 2\times 0.0005 = 1 \times 10^{-3} \, M }$
$ {pH = - log [H^+] = - log(1\times 10^{-3}) }$
$ {pH = 3 }$
$\because {pH +pOH = 14 }$
???$ {3 +pOH = 14 }$
????????$ { pOH = 14 - 3 }$
????????$ {pOH = 14 }$
????????$ {pOH = 11 }$
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- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
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