(D) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.Given $[H^{+}] = 3.981 \times 10^{-7} \ M$.Substituting the value: $p

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(D) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.Given $[H^{+}] = 3.981 \times 10^{-7} \ M$.Substituting the value: $pH = -\log (3.981 \times 10^{-7})$.Using the property $\log (a \times b) = \log a + \log b$: $pH = -(\log 3.981 + \log 10^{-7})$.$pH = -(\log 3.981 - 7)$.$pH = 7 - \log 3.981$.Given $\log 3.981 = 0.6000$,so $pH = 7 - 0.6000 = 6.4$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Water hydrolysis and pH scale

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

MHT CET 2022 All MHT CET

A
$5.6$
B
$4.6$
C
$7.6$
D
$6.4$

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2 Equilibrium: Ionic Equilibrium

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

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