Question

20 ml of 0.1 molar acetic acid is mixed with 50 molar of ch3 c o ok

Answer 1

Answer:

pKa = 4.74

pH = pKa + log[Salt]/[Acid]

4.8 = 4.74 + log([Salt] x 7)/(0.2)

0.06 = log([Salt] x 7)/(0.2)

Taking Antilogs,

1.148 = [Salt] x 7/(0.2)

Or [Salt] = (1.148 x 0.2)/7 = 0.04 M