(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 6-2.Equilibrium-II (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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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

What is the $pH$ value of $\frac{N}{1000} \ KOH$ solution?

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When $10 \ mL$ of $0.1 \ M \ HCl$ and $40 \ mL$ of $0.2 \ M \ H_2SO_4$ solutions are mixed,the $pH$ of the resulting mixture will be:

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How many litres of water must be added to $1 \ L$ of an aqueous solution of $HCl$ with a $pH$ of $1$ to create an aqueous solution with $pH$ of $2$?

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The $pH$ of a $10^{-3} \ M$ solution of hydrochloric acid will be:

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The $pH$ of the aqueous solution containing $0.49 \ g$ of $H_2SO_4$ in one litre is

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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

What is the $pH$ value of $\frac{N}{1000} \ KOH$ solution?

When $10 \ mL$ of $0.1 \ M \ HCl$ and $40 \ mL$ of $0.2 \ M \ H_2SO_4$ solutions are mixed,the $pH$ of the resulting mixture will be:

How many litres of water must be added to $1 \ L$ of an aqueous solution of $HCl$ with a $pH$ of $1$ to create an aqueous solution with $pH$ of $2$?

The $pH$ of a $10^{-3} \ M$ solution of hydrochloric acid will be:

The $pH$ of the aqueous solution containing $0.49 \ g$ of $H_2SO_4$ in one litre is

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