Question

What is the ph of a buffer that consists of 0.45 m ch3cooh and 0.35 m ch3coona? ka = 1.8 × 10–5?

Answer 1

4.64

Step-by-step explanation:

pKa= -logKa = -log (1.8 x10^-5)

pH = {pKa + log[salt/acid]}

=4.74 + log (0.35/0.45)

= 4.74 - 0.109

= 4.63

Answer 2

Answer is: pH of a buffer is 4.64.

ck(CH₃COOH) = 0.45 M.

cs(CH₃COONa) = 0.35 M.

Ka = 1.8·10⁻⁵.

pKa = -logKa.
pKa = -log(1.8·10⁻⁵) = 4.75.
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
pH = 4.75 + log(0.35M/0.45M).
pH = 4.75 - 0.11.
pH = 4.64.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.