Question

A 35.0 mL solution of 0.0834 M barbituric acid (HA, pKa = 1.01, Ka=9.8 x 10^-5) was titrated with 0.2500 M NaOH. Calculate the pH at the following volumes of added base

V_b = 0.00mL

V_b = 3.05mL

V_b = 1/2 V_equivalence

V_b = V_equivalence

V_b = 12.5 mL

Answer 1

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

M = moles / volume

pH = -log[H+]

a) HA -> H+ + A-

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

a) no NaOH

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

Assume [H+] = [A-] = x

[HA] = M – x

Substitute in Ka

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

Ka = x*x/(M-x)

9.8*10^-5 = x*x/(0.0834-x)

This is quadratic equation

x =0.00281

For pH

pH = -log(H+)

pH =-log(0.00281)

pH in a = 2.55

b) 3.05 ml NaOH

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

NOTE that in this case [A-] < [HA]; Expect pH lower than pKa

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 35*0.0834 = 2.919 mmol of acid

mmol of base = MV = 3.05*0.25 = 0.7625 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 2.919 -0.7625   = 2.1565 mmol

mmol of conjguate left = 0 + 0.7625 = 0.7625   

Get pKa

pKa = -log(Ka)

pKa = -log(9.8*10^-5) = 4.01

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.01+ log (0.7625 /2.1565 )

pH = 3.56

c) for half ml

in half equivalence point

pH = pKa

since

pH = pKa + log(A-/HA)

A- = HA (half implies 50% of ech so 50/50% = 1)

log(1) = 0

pH = pKa

pH = 4.01

d) Addition of Same quantitie of Acid/Base

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(9.8*10^-5) = 1.0204*10^-10

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

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

Assume [HA] = [OH-] = x

[A-] = M – x

Substitute in Kb

1.0204*10^-10 = [x^2]/[M-x]

recalculate M:

mmol of conjugate = 2.1565 mmol

Vrequired = mmol/M =(2.1565 )/(0.25) = 8.626 mL o fbase

Total V = V1+V2 = 35+8.626 = 43.626

[M] = 2.1565 /43.626= 0.0494 M

1.0204*10^-10 = [x^2]/[0.0494-x]

x =2.22*10^-6

[OH-] 2.22*10^-6

Get pOH

pOH = -log(OH-)

pOH = -log (2.22*10^-6) = 5.65

pH = 14-pOH = 14-5.65= 8.35

e) Addition of base

There will be finally an Excess of Base!

mol of acid < mol of base

Calculate pOH directly

[OH-] = M*V / Vt

mmol of acid = MV = 35*0.0834 = 2.919

mmol of base = MV = 0.25*12.5= 3.125

therefre,

mmol of strng base left = 3.125-2.919= 0.206 mmol

Vtotal = 35+12.5= 47.5 mL

[OH-] = 0.206 /47.5 = 0.0043

pOH = -log(OH-)

pOH = -log(0.0043) = 2.37

pH = 14-pOH = 14-2.37= 11.63

pH = 11.63