(A) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.Given,$[H^{+}] = 5.5 \times 10^{-3} \ mol \ L^{-1}$.Substituting the

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(A) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.Given,$[H^{+}] = 5.5 \times 10^{-3} \ mol \ L^{-1}$.Substituting the value into the formula:$pH = -\log (5.5 \times 10^{-3})$$pH = -(\log 5.5 + \log 10^{-3})$$pH = -(\log 5.5 - 3)$$pH = 3 - \log 5.5$Since $\log 5.5 \approx 0.74$,$pH = 3 - 0.74 = 2.26$.

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

Medium

If the hydrogen ion concentration of a given solution is $5.5 \times 10^{-3} \ mol \ L^{-1}$,the $pH$ of the solution will be

A
$2.26$
B
$3.4$
C
$3.75$
D
$2.76$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

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If the hydrogen ion concentration of a given solution is $5.5 \times 10^{-3} \ mol \ L^{-1}$,the $pH$ of the solution will be

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