A trace of base is added to pure water at 2 5 ∘ C so that [ H + ] falls to one-fourth of its original value. The pH of the solution is: ( lo g 2 = 0.3 )

Pure water: [ H + ] = 1 0 − 7 M . New [ H + ] = 2.5 × 1 0 − 8 M . pH = 7 + lo g 4 = 7 + 0.6 = 7.6 . Answer: (A).

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesEasy2 minQB

ChemistryEasysingle choice

A trace of base is added to pure water at $2 5^{\circ} C$ so that $[H^{+}]$ falls to one-fourth of its original value. The pH of the solution is: ( $lo g 2 = 0.3$ )

Options:

Answer:

Solution:

Pure water: $[H^{+}] = 1 0^{- 7} M$ .

New $[H^{+}] = 2.5 \times 1 0^{- 8} M$ .

$pH = 7 + lo g 4 = 7 + 0.6 = 7.6$ .

Answer: (A).

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

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