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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

the answer is c

tell me if wrong