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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumHydrolysis of SaltsMedium2 minPYQ_2023
ChemistryMediumstatement

20 mL of 0.1MNaOH is added to 50 mL of 0.1M acetic acid solution. The pH of the resulting solution is ×10-2. (Nearest integer) Given : pKaCH3COOH=4.76

log 2 = 0.30

log 3 = 0.48

Answer:
458
Solution:

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

pH=pKa+logsaltacid

CH3COOH+NaOHCH3COONa+H2O5                2                                         3                 0                 2

pH=pKa+log23

pH=4.76+0.30-0.48

= 4.76  .18

= 4.58

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Hydrolysis of Salts
2mℹ️ Source: PYQ_2023

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