(D) $KOH$ is a strong base, so $[OH^-] = [KOH] = 10^{-4} \ M$. $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$. At $25^\circ C$, $pH + pOH = 14$. $pH = 14 -

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(D) $KOH$ is a strong base, so $[OH^-] = [KOH] = 10^{-4} \ M$. $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$. At $25^\circ C$, $pH + pOH = 14$. $pH = 14 - 4 = 10$.

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

Medium

The $pH$ of $10^{-4} \ M$ $KOH$ solution will be

WBJEE 2014 All WBJEE

A
$4$
B
$11$
C
$10.5$
D
$10$

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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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The $pH$ of $10^{-4} \ M$ $KOH$ solution will be

Similar Questions

Which of the following solutions will have $pH$ close to $1.0$?

The $pH$ of $1 \ L$ of an aqueous solution of $HCl$ is $1$. How many liters of water must be added to make the $pH$ of this solution $2$ (in $L$)?

$1 \, mL$ of $0.1 \, M$ $HCl$ is added to $999 \, mL$ of $0.1 \, N$ $NaCl$ solution. The $pH$ of the resulting solution will be:

The $pH$ of a $0.05 \,M$ solution of a strong dibasic acid $(H_2A)$ is:

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