Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesMedium2 minPYQ_2022

ChemistryMediumsingle choice

$100 mL$ of $2 M$ of formic acid $(p K_{a} = 3.74)$ is neutralise by $NaOH$ , at the equivalence point $pH$ is

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Solution:

Sodium formate is present at the equivalence point. It is the salt of weak acid $+$ strong base. So, final solution will be basic in nature. As we know $pH = 7 + \frac{p K _{a}}{2} + \frac{l o g C}{2}$ where, $C$ is concentration of salt. Total volume of solution $= 100 + 100 = 200 mL$ Concentration of salt $(C) = 2 \times \frac{100}{200} = 1 M$ $pH = 7 + \frac{3.74}{2} + \frac{l o g [ 1 ]}{2}$ $pH = 7 + 1.87 + 0$ $pH = 8.87$

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

⏱ 2mℹ️ Source: PYQ_2022

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