Suppose that 25.0 mL of 0.100 M CH_3COOH (aq) is titrated with 0.100 M NaOH (aq)

Question

Suppose that 25.0 mL of 0.100 M CH_3COOH (aq) is titrated with 0.100 M NaOH (aq). For acetic acid, K_a = 2.62 times 10^-4. What is the initial pH of the 0.100 M CH_3COOH (aq) solution? What is the pH after the addition of 10.0 mL of 0.100 M NaOH (aq)? What volume of 0.100 M NaOH (aq) is required to reach halfway to the stoichiometric point? Please calculate the pH at that halfway point. What volume of 0.100 M NaOH (aq) is required to reach the stoichiometric point? Please calculate the pH at the stoichiometric point.

Explanation / Answer

a)

initially:

HA <-> H+ + A-

Ka = [H+][A-]/[HA]

2.62*10^-4 = x*X/(0.1-x)

x =[h+] = 0.00498

pH = -log(0.00498) = 2.30

b)

pH after

mmol of base = MV = 10*0.1 = 1 mmol of base

mmol of acid = MV = 25*0.1 = 2.5 mmol of acid

after reaction

mmol of acid left = 2.5 - 1 = 1.5 mmol

mmol of conjguate formed= 0 +1 = 1

this is a buffer

so

henderson hasselbach equation

pH = pKa + loG(A-/HA)

pKa = -log(2.62*10^-4) = 3.58169

pH = 3.58169 + log(1/1.5) =3.4055

c)

forequivalence:

mmol of acid = 2.5

so we need 2.45 mmol of b ase

M = mmol/mL

mL = mmol/M = 2.5/0.1 = 25 mL of base required...

but we need half way... so

V = 25/2 = 12.5 mL

this is a special cae sicne conjugate = acid

so

the buffer forms:

pH = pKa + l oG(a-/HA)

A-= HA = 1

so

pH = pKa + log(1)

pH = pKa = 3.58169

d)

as previosly calculated, V = 25 mL

then

A- + H2O <-> HA + Oh-

Kb = [HA ][OH-]/[A-]

10^-14)/(2.62*10^-4) = x*x/(0.05-x)

3.8167*10^-11 =  x*x/(0.05-x)

x = 1.38*10^-6

pOH = -log(x) = -log( 1.38*10^-6) = 5.8601

pH = 14- 5.8601= 8.1399

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