A1.36 l buffer solution consists of 0.109 m butanoic acid and 0.266 m sodium butanoate. calculate the ph of the solution following the addition of 0.067 moles of naoh . assume that any contribution of the naoh to the volume of the solution is negligible. the ka of butanoic acid is 1.52×10−5 .

the pH of the solution is 5.54

Explanation:

Butanoic acid = HA

n = mole

Sodium acetate = A

Ka = 1.52×10⁻⁵ thus pKa = -log of Ka = 4.82

n=C×V; n of HA=0.109×1.36=0.148 mole; n of A=0.266×1.36=0.362 mole

Adding 0.067 mole of NaOH, it reacts with HA and increase amount of A

After reaction, n of HA=0.148-0.067=0.081; n of A=0.362+0.067=0.429 mole

c=n/v and v=1.36L

using Hasselbach equation: pH = pKa+log[A]/[HA]

pH =4.82+log[(0.429÷1.36)/(0.081÷1.36)] = 5.54

the answer for sodium is 0.186

the answer for rubidium is 0.251

i'm pretty sure its b i might be wrong though!