(B) In an aqueous solution at $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.We know that $K_w = [H^+][OH^-] = 1.0 \times 10^{-

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(B) In an aqueous solution at $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.We know that $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.Taking the negative logarithm on both sides,we get $-\log K_w = -\log([H^+][OH^-]) = -\log(1.0 \times 10^{-14})$.This simplifies to $pK_w = pH + pOH = 14$.

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

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In an aqueous solution,the sum of $pH$ and $pOH$ is equal to ......

A
$7$
B
$14$
C
$0$
D
$1$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

The $pH$ of pure water or neutral solution at $50\,^{\circ}C$ is .... ($pK_w = 13.26$ at $50\,^{\circ}C$)

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What will be the sum of $pH$ and $pOH$ in an aqueous solution?

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At $90\,^{\circ}C$ pure water has $[H_3O^{+}] = 10^{-6}\,M$. The value of $K_w$ at this temperature will be:

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If the value of the dissociation constant of water is $1.8 \times 10^{-16}$,what is the ionic product of water?

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At $25^{\circ}C$,if $[OH^{-}] = 10^{-9} \ M$,find the $pH$ value of the solution.

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In an aqueous solution,the sum of $pH$ and $pOH$ is equal to ......

Similar Questions

The $pH$ of pure water or neutral solution at $50\,^{\circ}C$ is .... ($pK_w = 13.26$ at $50\,^{\circ}C$)

What will be the sum of $pH$ and $pOH$ in an aqueous solution?

At $90\,^{\circ}C$ pure water has $[H_3O^{+}] = 10^{-6}\,M$. The value of $K_w$ at this temperature will be:

If the value of the dissociation constant of water is $1.8 \times 10^{-16}$,what is the ionic product of water?

At $25^{\circ}C$,if $[OH^{-}] = 10^{-9} \ M$,find the $pH$ value of the solution.

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