A. Write the equilibrium reaction for C6H6COOH in water. B. i) write the neutral

Question

A. Write the equilibrium reaction for C6H6COOH in water. B. i) write the neutralization reaction between C6H6COOH and 0.20 M NaOH.
ii) what is the mole ratio for C6H6COOH and NaOH for complete neutralization?
iii) calculate the pH of the 35.00 mL of 0.155 M C6H6COOH before the addition of NAOH.
iv) what volume in liters of the 0.30 M NaOH is required for a compete neutralization of the 35.00 mL of 0.155 M C6H6COOH?
V) what is the pH of the resulting solution at the equivalence point?
7. Acid-Base Equilibrium Benzoate ion (C6HeCoo) is the conjugate base of benzoic acid (C6HocooH). The Ka for benzoic acid is- 6.3 x 105. a) Write the equilibrium reaction for C6HeCooH in water. What is the mole ratio for c6H6cooH and NaoH for complete neutralization? (i) (iii) Calculate the pH of the 35.00 mL of0.155 M C6H6COOH before the addition of NaOH. (iv) volume in liters of the 0.20 s required for a complete neutralization of the 35.00 mL of 155 M (v) What is the pH of the resulting solution at the equivalence point? (v) Calculate the pH ofthe solution after adding a total of 80.00 mL of 0.20 MNaoH to the 35.00 mL of 0.155 M C6H6COOH.

Explanation / Answer

a) Reaction,

C6H6COOH + H2O <==> C6H6COO- + H3O+

b) Neutralization reaction

(i) Equation,

C6H6COOH + NaOH <==> C6H6COONa + H2O

(ii) mole ratio for complete reaction = 1

(iii) pH before addition of NaOH

Ka = [C6H6COO-][H3O+]/[C6H6COOH]

6.5 x 10^-5 = x^2/0.155

x = [H3O+] = 3.174 x 10^-3 M

pH = -log[H3O+] = 2.50

(iv) Volume of NaOH needed for complete neutralization

= 0.155 M x 25 ml/0.2 M = 27.125 ml

(v) pH at equivalence point

[C6H6COO-] formed = 0.155 M x 35 ml/62.125 ml = 0.087 M

C6H6COO- + H2O <==> C6H6COOH + OH-

let x amojt has hydrolyzed

Kb = Kw/Ka = [C6H6COOH][OH-]/[C6H6COO-]

1 x 10^-14/6.5 x 10^-5 = x^2/0.087

x = [OH-] =3.66 x 10^-6 M

pOH = -log[OH-] = 5.43

pH = 14 - pOH = 8.57

(vi) 80 ml NaOH added

excess [OH-] = 0.2 M x (80-27.125) ml/115 ml = 0.078 M

pOH = -loh[OH-] = 1.10

pH = 14 - pOH = 12.89

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