Question

What volume of a 0.1048 M NaOH is needed to neutralize (get to the equivalence point) 50.00 mL of a 0.0876 M acetic acid solution? What is the pH at the equivalence point? Show your work please!

Answer 1

molarity of NaOH (M1) = .1048 M Volume of NaOH (V1) = ?

molarity of acetic acid (M2) = .0876 M   Volume of acetic acid (V2) = 50 mL

V1M1 = V2M2

V1 = V2M2/V1 = (.0876 x 50)/.1048 = 41.79 mL

At equivalence point some of the sodium acetate will be hydrolysed in water to form CH3COOH and OH^-. The amount of CH3COOH and OH^- formed as a result of hydrolysis are equal.

[CH3COOH] = [OH^-],

K_h = K_w/K_a = [CH3COOH] x [OH^-] / [CH3COO^-] = [OH^-]²/  [CH3COO^-]

Say [CH3COO^-] = c M

K_w/K_a = [OH^-]²/ c or   [OH^-] = (c.K_w/K_a)^1/2

or [H⁺] = [OH^-]/K_w = (K_w.K_a/c)^1/2

P^H = 1/2[ - logKw - logKa + logc] = 1/2 P^kw + 1/2P^ka + 1/2logc]

P^H = 7 + 1/2 x 4.75 + 1/2 x log 0.0876 = 8.85