(A) For a strong acid like $HCl$,it dissociates completely in water: $HCl \rightarrow H^{+} + Cl^{-}$.Given concentration of $HCl = 0.10 \ M$,so $[H^{

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(A) For a strong acid like $HCl$,it dissociates completely in water: $HCl \rightarrow H^{+} + Cl^{-}$.Given concentration of $HCl = 0.10 \ M$,so $[H^{+}] = 0.10 \ M = 10^{-1} \ M$.The $pH$ of the solution is calculated as: $pH = -\log[H^{+}] = -\log(10^{-1}) = 1$.Using the relation at $25^{\circ}C$: $pH + pOH = 14$.Therefore,$pOH = 14 - pH = 14 - 1 = 13$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of strong Acids and strong Bases

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Calculate the $pOH$ of $0.10 \ M \ HCl$ solution.

AP EAMCET 2021 All AP EAMCET

A
$13$
B
$1$
C
$7$
D
$14$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

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Calculate the $pOH$ of $0.10 \ M \ HCl$ solution.

Similar Questions

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