(NONE) Given,$[OH^{-}] = 0.1 \ M = 10^{-1} \ M$.$pOH = -\log[OH^{-}] = -\log(10^{-1}) = 1$.Given the ionic product of water $K_w = [H^{+}][OH^{-}] = 1

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(NONE) Given,$[OH^{-}] = 0.1 \ M = 10^{-1} \ M$.$pOH = -\log[OH^{-}] = -\log(10^{-1}) = 1$.Given the ionic product of water $K_w = [H^{+}][OH^{-}] = 10^{-15}$,we have $pH + pOH = pK_w = 15$.Therefore,$pH = 15 - pOH = 15 - 1 = 14$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium pH of strong Acids and strong Bases

Medium

The $pH$ value of $0.1 \ M \ NaOH$ solution is (given that the ionic product of water $K_w = [H^{+}] [OH^{-}] = 10^{-15}$)

A
$13$
B
$12$
C
$11$
D
$2$

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2 Equilibrium: Ionic Equilibrium

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pH of strong Acids and strong Bases

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MediumTS EAMCET 2020

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The $pH$ value of $0.1 \ M \ NaOH$ solution is (given that the ionic product of water $K_w = [H^{+}] [OH^{-}] = 10^{-15}$)

Similar Questions

Number of equivalents of $H_2SO_4$ present in $100 \ mL$ of its solution whose $pH$ is $5$.

What will be the $pH$ of a solution formed by mixing $40 \, mL$ of $0.10 \, M \, HCl$ with $10 \, mL$ of $0.45 \, M \, NaOH$?

What is the concentration of hydrogen $[H^{+}]$ ions in $0.01 \, M \, HCl$?

$0.1 \ mL$ of $0.001 \ M$ $HCl$ solution is diluted with water to form $10 \ L$. Calculate the $pH$ of the dilute solution.

What is the $pH$ of a $10^{-8} \ M$ solution of $HCl$?

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