A0.100 m solution of butanoic acid has a % deprotonation of 1.23%. determine the ph of the butanoic acid solution.

Chemical reaction: C₃H₇COOH → C₃H₇COO⁻ + H⁺.
c(butanoic acid) = 0,100 M.
α = 1,23% = 0,0123.
Ka = α² · c / 1 - α.
Ka = 0,0123² · 0,1 M / 1 - 0,0123.
Ka = 0,0000153 M.
Ka = c(C₃H₇COO⁻) · c(H⁺) / c(C₃H₇COOH).
c(H⁺) = α · c(C₃H₇COOH).
c(H⁺) = 0,0123 · 0,1 M = 0,00123 mol/L.
pH = -log c(H⁺).
pH = 2,91.

Is a logical series of steps

Ireally don’t know this