Question

a solution is made by combining 15.0 mlml of 18.5 mm acetic acid with 5.54 gg of sodium acetate and diluting to a total volume of 1.50 ll.

Answer 1

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

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