(C) The formula for $pH$ is given by $pH = -\log [H^{+}]$.Given $[H^{+}] = 6 \times 10^{-4} \ M$.$pH = -\log (6 \times 10^{-4}) = -(\log 6 + \log 10^{

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(C) The formula for $pH$ is given by $pH = -\log [H^{+}]$.Given $[H^{+}] = 6 \times 10^{-4} \ M$.$pH = -\log (6 \times 10^{-4}) = -(\log 6 + \log 10^{-4})$.$pH = -(\log 2 + \log 3 - 4) = -(0.301 + 0.477 - 4) = -(-3.222) = 3.222$.Rounding to two decimal places,the $pH$ is $3.22$.

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

Medium

The hydrogen ion concentration in a given solution is $6 \times 10^{-4} \ M$. Its $pH$ will be

A
$6$
B
$4$
C
$3.22$
D
$2$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

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

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The $[OH^{-}]$ in a solution is $1 \times 10^{-12} \ mol \ dm^{-3}$. What is the concentration of $H^{+}$ ions?

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$100 \ mL$ of $NaOH$ solution has a $pH = 10$. Calculate the concentration of hydroxide ions,$[OH^{-}]$.

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$A$ $pH$ of $7$ signifies

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Derive the relation between $pH$ and $pOH$.

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The hydrogen ion concentration in a given solution is $6 \times 10^{-4} \ M$. Its $pH$ will be

Similar Questions

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

The $[OH^{-}]$ in a solution is $1 \times 10^{-12} \ mol \ dm^{-3}$. What is the concentration of $H^{+}$ ions?

$100 \ mL$ of $NaOH$ solution has a $pH = 10$. Calculate the concentration of hydroxide ions,$[OH^{-}]$.

$A$ $pH$ of $7$ signifies

Derive the relation between $pH$ and $pOH$.

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