Calculate the pOH of a solution 2 5 ∘ C that contains 1 × 1 0 − 10 M of hydronium ions i.e. H 3 O + :NEET 2007

Given [ H 3 O + ] = 1 × 1 0 − 10 M at 2 5 ∘ [ H 3 O + ] [ OH − ] = 1 0 − 14 ∴ [ OH − ] = 1 0 − 10 1 0 − 14 = 1 0 − 4 Now, [ OH − ] = 1 0 − p OH = 1 0 − 4 = 1 0 − P OH ∴ p OH = 4 .

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ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesMedium2 minPYQ_2007

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Calculate the $pOH$ of a solution $2 5^{\circ} C$ that contains $1 \times 1 0^{- 10} M$ of hydronium ions i.e. $H_{3} O^{+}$ :NEET 2007

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Solution:

Given $[H_{3} O^{+}] = 1 \times 1 0^{- 10} M$ at $2 5^{\circ} [H_{3} O^{+}] [OH^{-}] = 1 0^{- 14}$ $∴ [OH^{-}] = \frac{1 0 ^{- 14}}{1 0 ^{- 10}} = 1 0^{- 4}$ Now, $[OH^{-}] = 1 0^{- p^{OH}}$ $= 1 0^{- 4} = 1 0^{- P^{OH}}$ $∴ p^{OH} = 4$ .

Stream:NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Strong and Weak Acids and Bases

⏱ 2mℹ️ Source: PYQ_2007

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