Question

What is the pH at the equivalence point when 50.00 mL of 0.112 M hydroxyacetic acid...

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 Ka = 1.47???10?4, and Kw = 1.01???10?14.)

Homework Answers

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-

Kb = [(OH)CH2COOH][OH-] / [(OH)CH2COOH]

Kb = Kw/Ka

Kb = 1.01×10-14/1.47 ×10-4

Kb = 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

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