(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 2 Equilibrium: 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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Class 11

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Chapter

2 Equilibrium: Ionic Equilibrium

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Water hydrolysis and pH scale

Find the concentration of $(OH^-)$ in a solution with $pH = 3.3$.

Easy

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If the $pH$ of a solution is $4.0$ at $25\,^{\circ}C,$ its $pOH$ would be $(K_w = 10^{-14})$

Medium

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$A$ is an aqueous acid; $B$ is an aqueous base. They are diluted separately,then:

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At $100\, ^oC$,the $K_w$ of water is $55$ times its value at $25\, ^oC$. What will be the $pH$ of a neutral solution at this temperature? $(\log 55 = 1.74)$

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The concentration of hydrogen ions in a given solution is $6 \times 10^{-4} \ M$. The $pH$ of the solution is:

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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

Similar Questions

Find the concentration of $(OH^-)$ in a solution with $pH = 3.3$.

If the $pH$ of a solution is $4.0$ at $25\,^{\circ}C,$ its $pOH$ would be $(K_w = 10^{-14})$

$A$ is an aqueous acid; $B$ is an aqueous base. They are diluted separately,then:

At $100\, ^oC$,the $K_w$ of water is $55$ times its value at $25\, ^oC$. What will be the $pH$ of a neutral solution at this temperature? $(\log 55 = 1.74)$

The concentration of hydrogen ions in a given solution is $6 \times 10^{-4} \ M$. The $pH$ of the solution is:

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