Exams › NEET › Chemistry

The pH of solution containing \( 0.10 M \) sodium acetate and \( 0.03 M \) acetic acid is: \( \left(p K_{a} \text { for } C H_{3} C O O H=4.57\right) \)

1. 4.09
2. 6.09
3. 5.09
4. 7.09

Correct answer: 5.09

Solution

This is an acetic acid/acetate buffer, so the Henderson–Hasselbalch equation applies: pH = pKa + log([A−]/[HA]). Since acetate is the conjugate base and is more concentrated than the acid, the pH should be slightly above 4.57, matching 5.09.

Related NEET Chemistry questions

• 18 g glucose \( \left(C_{6} H_{12} O_{6}\right) \) is added to \( 178.2 \mathrm{g} \) of water. The vapour pressure of this aqueous solution at \( 100^{0} C \) in torr is:
• Distribution law was given by:
• Assertion The specific charge of canal rays is not constant. Reason The mass of ion varies with speed.
• Which is the condition, unrelated to quantities, that indicates that the rate going into solution is equal to the rate coming out of solution?
• For the equilibrium, \( \boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{s}) \Leftrightarrow \boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{l}) \) which of the following is true?
• The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by \( 0.45^{\circ} \mathrm{C} \). The degree of association of acetic acid in benzene is: [Assume acetic acid dimerizes in benzene and \( K_{f} \) for benzene \( =\mathbf{5 . 1 2} \mathbf{K} \) \( \left.\mathrm{kg} \mathrm{mol}^{-1}\right] \)

⚔️ Practice NEET Chemistry free + battle 1v1 →