Question

What is the pH when 5.3 g of sodium acetate, NaC2H3O2, is dissolved in 230.0 mL of water? (The Ka of acetic acid, HC2H3O2, is 1.8×10−5 .)

Answer 1

• CH₃COONa + HOH$\iff$ NaOH + CH₃COOH

CH₃COO^- + Na⁺ + HOH $\iff$ Na⁺ + OH^- + CH₃COOH

CH₃COO^- + HOH $\iff$ OH^- + CH₃COOH, so OH^--ion responsibles for alkalinity $\implies$ this salt has an alkaline environment.

• Calculation salt`s concentration: ν(CH₃COONa) = $m / M$ = 5.3 (g) / 82.03 (g/mole) = 0.06 mole;

C(CH₃COONa) = $ν/V$ = 0.06 (mole) / 0.23 (L) = 0.26 M

• Calculation pH: pH = $7+1/2(pKa)+1/2(lgC(salt))$ = 7 + 1/2(-lg1.8∙10^-5) + 1/2(lg0.26) = 7 + 2.37 - 0.3 = 9.07
• Answer: pH=9.07.