A 25.0 mL sample of a 0.100 M solution of acetic acid is titrated with a 0.125 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76*10-5.

Ater adding 10.0 mL the pH = 4.74

After adding 20.0 mL the pH = 8.75

After adding 30.0 mL the pH = 12.36

Explanation:

Step 1: Data given

Volume of a 0.100 M acetic acid solution = 25.0 mL

Molarity of NaOH solution = 0.125 M

A) 25.0mL of HAc (acetic acid) and 10.0mL of NaOH

Calculate moles

Equation: NaOH + HAc - -> NaAc + H20

Moles HAc = 0.025 L * 0.100 mol / L = 0.00250 moles

Moles NaOH = 0.01 L * 0.125mol/L = 0.00125 moles

Initial moles

NaOH = 0.00125 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00125 - 0.00125 = 0 moles

HAc = 0.00250 - 0.00125 = 0.00125 moles

NaAC = 0.00125 moles

Calculate molarity

[HAc] = [Ac-] = 0.00125 moles / 0.035 L = 0.0357M

Concentration at equilibrium

HAc Ac - + H + Ka = 1.76 * 10^-5

[HAc] = 0.0357 - x M

[Ac-] = 0.0357 + X M

[H+] = XM

Ka=1.76*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]

Ka=1.76*10^-5 = [0.00357][x]/[0.0357] usually good approx.

Ka = 1.76 * 10-5 = x = [H+]

pH = - log [H+] = - log [1.76*10^-5] = 4.74

B) 25.0mL HAc and 20.0mL NaOH

Calculate moles

Equation: NaOH + HAc - -> NaAc + H20

Moles HAc = 0.025 L * 0.100 mol / L = 0.00250 moles

Moles NaOH = 0.02 L * 0.125mol/L = 0.00250 moles

Initial moles

NaOH = 0.00250 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00250 - 0.00250 = 0 moles

HAc = 0.00250 - 0.00250 = 0 moles

NaAC = 0.00250 moles

Calculate molarity

[NaAC] = 0.00250 moles / 0.045 L = 0.0556M

Concentration at equilibrium

Ac - + H20 HAc + OH - Ka = 1.76 * 10^-5

[Ac] = 0.0556 - x M

[OH-] = XM

[HAc] = XM

**Kw = 1*10^-14 at STP I'll assume this for this problem

**Kw = Ka*Kb ⇒ Kb = Kw/Ka

**Kb = (1*10^-14) / (1.76*10^-5) = 5.68*10^-10

Kb = ([HAc][OH-]) / [Ac-] = ([x][x]) / [0.0556]

Kb = x² / 0.0556

5.68*10^-10 = x² / 0.0556

so x = 5.62*10^-6

pOH = - log [OH-] = 5.250

pH = 14 - pOH

pH 14 - 5.25 = 8.75

C) 25.0mL HAc and 30.0mL NaOH

NaOH + HAc - -> NaAc + H20

Initial numbers of moles

Moles NaOH = 0.125M * 0.03 L = 0.00375 moles

Moles HAc = 0.00250 moles

moles NaAC = 0 moles

Moles at the equilibrium

Moles NaOH = 0.00375 moles - 0.00250 = 0.00125 moles

Moles HAc = 0.00250 moles - 0.00250 = 0 moles

moles NaAC = 0.00250 moles

After the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.

Calculate NaOH molarity

(0.00125moles/0.055 L = 0.0227M NaOH

Strong Bases dissociate 100% so [OH-] = 0.0227M

pOH = - log[0.0227] = 1.644

pH = 14-pOH = 12.36

10mil 76 the experllon is the most numbers