JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 18)
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 $${1 \over 2}$$ 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.
Answer
5.22TO5.24
Explanation
milimole of acetic acid in 20 ml
= $${3 \over {60}} \times 1000 \times {{20} \over {500}}$$ = 2
milimole of HCl in 20 ml = 25 $$ \times $$ $${{20} \over {500}}$$ = 1
milimole of NaOH 20 ml = $${1 \over 2} \times 5$$ = 2.5
pH = pKa + log$${{1.5} \over {0.5}}$$
= 4.74 + log 3 = 5.22
= $${3 \over {60}} \times 1000 \times {{20} \over {500}}$$ = 2
milimole of HCl in 20 ml = 25 $$ \times $$ $${{20} \over {500}}$$ = 1
milimole of NaOH 20 ml = $${1 \over 2} \times 5$$ = 2.5
| NaOH | + | CH3COOH | $$ \to $$ | CH3COONa | + | H2O |
|---|---|---|---|---|---|---|
| 1.5 | 2 | 0 | 0 | |||
| 0 | 0.5 | 1.5 |
pH = pKa + log$${{1.5} \over {0.5}}$$
= 4.74 + log 3 = 5.22
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