(D) $H_2SO_4 \longrightarrow 2H^{+} + SO_4^{2-}$ $[H_2SO_4] = 0.005 \ M$ $[H^{+}] = 2 \times 0.005 = 0.01 \ M$ $pH = -\log[H^{+}] = -\log(0.01)$ $pH =

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(D) $H_2SO_4 \longrightarrow 2H^{+} + SO_4^{2-}$ $[H_2SO_4] = 0.005 \ M$ $[H^{+}] = 2 \times 0.005 = 0.01 \ M$ $pH = -\log[H^{+}] = -\log(0.01)$ $pH = 2$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of strong Acids and strong Bases

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What is the $pH$ of $0.005 \ M \ H_2SO_4$ solution?

MHT CET 2021 All MHT CET

A
$5$
B
$2.3$
C
$3.3$
D
$2$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

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What is the $pH$ of $0.005 \ M \ H_2SO_4$ solution?

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What is the $pH$ of a $2.6 \times 10^{-8} \ M \ H^{+}$ ion solution?$(\log 2.6 = 0.4150)$

When $10^{-8} \ mol$ of $HCl$ is dissolved in one litre of water,the $pH$ of the solution will be

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What is the $pH$ of a $2.6 \times 10^{-8} \ M \ H^{+}$ ion solution?
$(\log 2.6 = 0.4150)$