JEE Main 2020 — Ionic Equilibrium Question with Solution
From: JEE Main 2020 (Online) 7th January Evening Slot
Question
3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL
of this solutions mL of 5 M NaOH is added. The pH of the solution is __________.
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.
Enter your answer
Show full solutionCorrect answer: 5.22
Correct answer
5.22
Step-by-step explanation
milimole of acetic acid in 20 ml
= = 2
milimole of HCl in 20 ml = 25 = 1
milimole of NaOH 20 ml = = 2.5
pH = pKa + log
= 4.74 + log 3 = 5.22
= = 2
milimole of HCl in 20 ml = 25 = 1
milimole of NaOH 20 ml = = 2.5
| NaOH | + | CH3COOH | CH3COONa | + | H2O | |
|---|---|---|---|---|---|---|
| 1.5 | 2 | 0 | 0 | |||
| 0 | 0.5 | 1.5 |
pH = pKa + log
= 4.74 + log 3 = 5.22
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