At 10 0 ∘ C the K w of water is 55 times its value at 2 5 ∘ C . What will be the pH of neutral solution? ( lo g 55 = 1.74 ) NEET 2013 (Karnataka)

H + pH pH = 55 × 1 0 − 14 = − lo g [ H + ] = − lo g [ 55 × 1 0 − 14 ] = 2 1 [ − lo g ( 55 × 1 0 − 14 ) ] = 6.13

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesHard2 minPYQ_2013

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At $10 0^{\circ} C$ the $K_{w}$ of water is 55 times its value at $2 5^{\circ} C$ . What will be the $pH$ of neutral solution? $(lo g 55 = 1.74)$ NEET 2013 (Karnataka)

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Answer:

Solution:

$H^{+} pH pH = 55 \times 1 0^{- 14} = - lo g [H^{+}] = - lo g [55 \times 1 0^{- 14}] = \frac{1}{2} [- lo g (55 \times 1 0^{- 14})] = 6.13$

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

⏱ 2mℹ️ Source: PYQ_2013

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