(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

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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

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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

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$0.001 \ mol$ of a strong electrolyte $M(OH)_2$ is dissolved to make $20 \ mL$ of a saturated solution. Calculate its $pH$ $(K_w = 1 \times 10^{-14})$.

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