Q.55. 20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A so

Q: What is the correct answer for JEE Main Chemistry Q55?

The correct answer is B. Sol. For neutralization of acetic acid: $M_{CH_3COOH} V_{CH_3COOH} = M_{NaOH} V_{NaOH}$ $M_{CH_3COOH} \times 20 = 0.1 \times 28.4$ $M_{CH_3COOH} = \frac{0.1 \times 28.4}{20} = 0.142$ M In solution (X): Moles of $CH_3COOH = 0.142 \text{ M} \times 20 \text{ mL} = 2.84 \text{ mmol}$ Moles of $NaOH = 0.1 \text{ M} \times 14.2 \text{ mL} = 1.42 \text{ mmol}$ Reaction: $CH_3COOH + NaOH \rightarrow CH_3C

Q: Q.55. 20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH…

The correct answer is B. Sol. For neutralization of acetic acid: $M_{CH_3COOH} V_{CH_3COOH} = M_{NaOH} V_{NaOH}$ $M_{CH_3COOH} \times 20 = 0.1 \times 28.4$ $M_{CH_3COOH} = \frac{0.1 \times 28.4}{20} = 0.142$ M In solution (X): Moles of $CH_3COOH = 0.142 \text{ M} \times 20 \text{ mL} = 2.84 \text{ mmol}$ Moles of $NaOH = 0.1 \text{ M} \times 14.2 \text{ mL} = 1.42 \text{ mmol}$ Reaction: $CH_3COOH + NaOH \rightarrow CH_3C

Question

Q.55. 20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X) ? (pK value of acetic acid is 4.75).

FirstInTest · Accepted Answer

Sol. For neutralization of acetic acid: $M_{CH_3COOH} V_{CH_3COOH} = M_{NaOH} V_{NaOH}$ $M_{CH_3COOH} 	imes 20 = 0.1 	imes 28.4$ $M_{CH_3COOH} = \frac{0.1 	imes 28.4}{20} = 0.142$ M In solution (X): Moles of $CH_3COOH = 0.142 	ext{ M} 	imes 20 	ext{ mL} = 2.84 	ext{ mmol}$ Moles of $NaOH = 0.1 	ext{ M} 	imes 14.2 	ext{ mL} = 1.42 	ext{ mmol}$ Reaction: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$ Initial: 2.84 mmol 1.42 mmol 0 After reaction: $(2.84 - 1.42)$ mmol 0 1.42 mmol So, $CH_3COOH = 1.42 	ext{ mmol}$ and $CH_3COONa = 1.42 	ext{ mmol}$ This is a buffer solution with [acid] = [salt]. Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ $pH = 4.75 + \log \frac{1.42}{1.42}$ $pH = 4.75 + \log 1$ $pH = 4.75 + 0 = 4.75$

Answer

0

Answer

45

Answer

5

Answer

82

Related Questionsin Chemistry

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions