pH of the solution is 4.14.
m(CH₃COONa) = 5.54 g.
n(CH₃COONa) = m(CH₃COONa) ÷ M(CH₃COONa).
n(CH₃COONa) = 5.54 g ÷ 82.034 g/mol.
n(CH₃COONa) = 0.067 mol; amount of sodium acetate.
V(CH₃COONa) = 1.5 L
c(CH₃COONa) = n(CH₃COONa) ÷ V(CH₃COONa).
c(CH₃COONa) = 0.067 mol ÷ 1.5 L.
c(CH₃COONa) = 0.045 M; molarity of sodium acetate in solution.
c(CH₃COOH) = 18.5 M; molarity of pure acetic acid.
V(CH₃COOH) = 15.0 mL = 0.015 L; volume of acetic acid
n(CH₃COOH) = 18.5 M x 0.015 L = 0.2775 mol; amount of pure acetic acid
c(CH₃COOH) = 0.2775 mol / 1.5 L = 0.185 M; concentration of acetic acid in solution
Ka(CH₃COOH) = 1.8·10⁻⁵
pKa = -logKa = 4.75
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
pH = 4.75 + log(0.045M / 0.185M)
pH = 4.14; potential of hydrogen
More about sodium acetate: brainly.com/question/24671704
#SPJ4
