Question

#21

Calculate the pH of a solution prepared by mixing 49.0 mL of butyric acid, HC₄H₇O₂ , with 6.15g of KOH in water. The following data about butyric acid may be helpful:

density = 0.9595 g/mol; K_a = 1.54 x 10^-5

Answer 1

MW of acid = 88.1051

mol of acid = mass/MW

mass of acid = D*V = 49*0.9595 = 47.0155

mol = 47.0155/(88.1051) = 0.53362972177 mol of acid

now

mol of KOH = mass/MW = 6.15/56.1056 = 0.109614 mol of acid

Total volume = 49 ml (since no other data is stated)

therefore

mol of acid left = 0.53363-0.109614 = 0.424016

mol of base formed = 0.109614

this is a buffer so

pH = pKa + log(base/acid)

pKa = -log(ka) = -log(1.54*10^-5) = 4.81

pH = 4.81 + log(0.109614/0.424016) = 4.2224837

pH = 4.2224837