Question

If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration

Answer 1

CH₃CH₂COOH + H₂O ↔ CH₃CH₂COO⁻ + H₃O⁺

pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration