A solution is prepared by mixing 2.00 g propionic acid with 0.45g of solid NaOH

Question

A solution is prepared by mixing 2.00 g propionic acid with 0.45g of solid NaOH in water to make 500.0 mL

7. A solution is prepared by mixing 2.00 g propionic acid, HC,HO: (MW74 g/mol, K 1.3 x 10) with 0.45 g of solid NaOH (MW-40 g/mol) in water to make 500.0 ml. (2.0 pts) a) Write the complete chemical equation including phases for the neutralization reaction that occurs. Use on of the following amows: or , to show the equilibrium of the reaction. -or b) Write any electrolytes in the above neutralization products in their dissociated form. c) Calculate moles of NaOH added: d) Calculate moles of propionic acid added: e) Determine the pH of the resulted solution?

Explanation / Answer

a) CH3CH2-COOH(aq) + NaOH(aq) ---> CH3CH2-COONa(aq) + H2O(l)

b) CH3CH2-COOH(aq) ----> CH3CH2-COO^-(aq) + H^+(aq)

      NaOH(aq) ---> Na^+(aq) + OH^-(aq)

CH3CH2-COONa(aq) ---> CH3CH2-COO^-(aq) + Na^+(aq)

c) no of mol of NaOH = w/Mwt = 0.45/40 = 0.01125 mol

d) no of mol of CH3-CH2-COOH = 2/74 = 0.027 mol

E) pka of CH3-CH2-COOH = -logka = -log(1.3*10^-5) = 4.89

     pH = pka +log(salt/acid)

        = 4.89+log(0.01125/(0.027-0.01125))

        = 4.744

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