(D) The ionic product of water is given by $K_{w} = [H^{+}][OH^{-}] = 10^{-14} \ M^{2}$ at $298 \ K$. Given $[H^{+}] = 0.001 \ M = 10^{-3} \ M$. There

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(D) The ionic product of water is given by $K_{w} = [H^{+}][OH^{-}] = 10^{-14} \ M^{2}$ at $298 \ K$. Given $[H^{+}] = 0.001 \ M = 10^{-3} \ M$. Therefore,$[OH^{-}] = \frac{K_{w}}{[H^{+}]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ M$.

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

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$A$ solution has $[H^{+}] = 0.001 \ M$. What is the value of $[OH^{-}]$?

MHT CET 2022 All MHT CET

A
$10^{-2} \ M$
B
$10^{-3} \ M$
C
$1 \ M$
D
$10^{-11} \ M$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

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$A$ solution has $[H^{+}] = 0.001 \ M$. What is the value of $[OH^{-}]$?

Similar Questions

What is the $pH$ of a solution having $H^{+}$ ion concentration $3.981 \times 10^{-7} \ M$? $(\log 3.981 = 0.6000)$

The concentration of hydronium $({H_3}{O^{+}})$ ion in pure water at $298 \ K$ is:

The $pH$ of neutral water at room temperature is nearly:

For the reaction $B^{-} + H_2O \rightleftharpoons HA + OH^{-}$,the value of the hydrolysis constant at equilibrium is equal to $K_a^{-1} = \dots \dots$

$A$ solution whose $pH$ value is less than $7$ will be

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