Question

What is the pH at the equivalence point when 50.00 mL of 0.112 M hydroxyacetic acid is titrated with 0.0580 M KOH? (Assume K_a = 1.47???10^?4, and K_w = 1.01???10^?14.)

Answer 1

(OH)CH2COOH + OH^- ---------> (OH)CH2COO^- + H2O

1:1molar reaction

M1V1 = M2V2

V2 = M1V1/M2

= 50.00mL × 0.112M/ 0.0580M

= 96.55mL

At equivalence point ,

Total volume = 146.55ml

[(OH)CH2COO^-] = 0.112M / ( 146.55ml / 96.55ml)

[(OH)CH2COO^-] = 0.07379M

(OH)CH2COO^- partly hydrolysed by water

(OH)CH2COO^- + H2O <--------> (OH)CH2COOH + OH^-

K_b = [(OH)CH2COOH][OH^-] / [(OH)CH2COOH]

K_b = K_w/K_a

K_b = 1.01×10^-14/1.47 ×10^-4

K_b = 6.87 × 10^-11

At equilibrium

[(OH)CH2COO^-] = 0.07379 - x

[(OH)CH2COOH] = x

[OH^-] = x

so,

x²/( 0.07379 - x) = 6.87 ×10^-11

we can assume , 0.07379 - x = 0.07379 because x is small value

x²/ 0.07379 = 6.87 ×10^-11

x² = 5.07 × 10^-12

x = 2.25 × 10^-6

[OH^-] = 2.25 ×10^-6M

pOH = -log[OH^-]

pOH = - log( 2.25 ×10^-6M)

pOH = 5.65

pH = 14 - 5.65

pH = 8.35