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
