What will be the pH of a solution containing an acid of pKa1.9, with an acid concentration twenty times that of the conjugate base?

Question

What will be the pH of a solution containing an acid of pKa1.9, with an acid concentration twenty times that of the conjugate base?

Answer 1

Assuming the acid is of the form HA, then the conjugate base is A- and using the Henderson-Hasselbalch equation we have pH = pKa + log [A-]/[HA]

Since [HA] is 20 time [A-], this becomes pH = pKa + log (1/20) = pKa - 1.3

pH = 1.9 - 1.3 = 0.6