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.

Dot structures do not show the distribution of electrons in orbitals and take up a lot of space.

Arrow and line diagrams take up a lot of space and make it difficult to count electrons.

Written configurations make it easy to lose count of electrons and do not show the distribution of electrons in orbitals.

(re-write in your own words this is off engenuity)