(B) The ionic product of water is given by the expression: $K_w = [H_3O^{+}][OH^{-}]$.For pure water,the concentration of hydronium ions is equal to t

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(B) The ionic product of water is given by the expression: $K_w = [H_3O^{+}][OH^{-}]$.For pure water,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 10^{-6}\, mol\, L^{-1}$.Substituting these values into the expression for $K_w$:$K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Water hydrolysis and pH scale

Medium

At $90\, ^\circ C$ pure water has $[H_3O^{+}] = 10^{-6}\, mol\, L^{-1}$. The value of $K_w$ at $90\, ^\circ C$ is

A
$10^{-6}$
B
$10^{-12}$
C
$10^{-14}$
D
$10^{-8}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

Find the concentration of hydrogen ions in a solution with $pH = 5.4$ in $mol/L$.

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At $90 \, ^\circ C$,the concentration of $[H_3O^+]$ in pure water is $10^{-6} \, mol \, L^{-1}$. What is the value of $K_W$ at $90 \, ^\circ C$?

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On adding which of the following,the $pH$ of $20 \ mL$ of $0.1 \ N \ HCl$ will not alter?

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For an acidic solution,the $[OH^{-}]$ will be .........

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At $90\, ^\circ C$ pure water has $[H_3O^{+}] = 10^{-6}\, mol\, L^{-1}$. The value of $K_w$ at $90\, ^\circ C$ is

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Find the concentration of hydrogen ions in a solution with $pH = 5.4$ in $mol/L$.

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At $90 \, ^\circ C$,the concentration of $[H_3O^+]$ in pure water is $10^{-6} \, mol \, L^{-1}$. What is the value of $K_W$ at $90 \, ^\circ C$?

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