Question

One litre of an aqueius solution contain 0.15 mole of ch3cooh pka =4.8and 0.15mole of ch3coona after the addition of 0.05 mole of solid naoh to this the ph will be

Answer 1

Answer:

5.1

Explanation:

ph = pka + log [conjugate base/acid ]

Answer 2

The pH will be 5.1

pH of a acidic buffer can be calculated by using Handerson -Hasselbach equation-

pH=pKa + log* [conjugate base]/[Acid]

here, by adding .05 mole of NaoH the concentration of CH3COOH will increase due to common ion effect and the concentration of [H+] will decrease and hence the concentration of conjugate base will be 0.2 moles and that of acid will be 0.1.

Substituting these values in the above equation you will get the answer.

*the log scale in base 10