help with this chem question please?

Question

Find the pH and volume (mL) of 0.0372 M NaOH needed to reach the equivalence point(s) in titrations of the following.

(a) 47.2 mL of 0.0526 M CH3COOH

(b) 17.6 mL of 0.0858 M H2SO3 (two equivalence points)

Answer 1

You have weak acids so the salts formed at the equivalence point will hydrolyse.

(a) CH3COOH + NaOH -> CH3COONa + H2O

The stoichiometry is 1:1 so at the equivalence point
mole acid =mole base =>
Ca*Va = Cb*Vb => Vb= (Ca/Cb)*Va =(0.0526/0.0372)*47.2 = 66.7 ml NaOH.

Also the stoichiometry is 1:1 for the production of CH3COONa from CH3COOH, so since all the moles of the acid have reacted, you have
moles CH3COONa = moles CH3COOH = Ca*Va
But since you added base the volume is now V=Va+Vb
So the concentration of CH3COONa is
C=mole/V= Ca*Va/(Va+Vb)= 0.0526*47.2/(47.2+66.7) = 0.0218 M

.. .. .. .. .. .. CH3COO(-) + H2O <=> CH3COOH + OH(-)
Initial .. .. .. .. .. C
React .. .. .. .. . x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
At Equil. .. .. ..C-x .. .. .. .. .. .. .. .. .. .. .. .. .x .. .. .. .. .. x

Kb=[CH3COOH][OH-]/[CH3COO-]= x^2/(C-x)
Also Kb=Kw/Ka and Ka=10^-pKa
For acetic acid pKa=4.76, we found that C=0.0218, so by sustituting we get
10^-14/10^-4.76=x^2/(0.0218-x)
Let's assume that 0.0218>> x and 0.0218-x=0.0218.
Then the equation becomes 10^-9.24= x^2/0.0218 =>
x=3.54*10^-6 << 0.0218 so our assumption is OK
pH=14-pOH=14-(-logx) =14+log(3.54*10^-6) = 8.55

(b) For the first equivalent point
H2SO3 + NaOH -> NaHSO3 + H2O
It is exactly the same as above; only the values change.
The equilibrium reaction will be
HSO3(-) + H2O <=> H2SO3 +OH(-)
with Kb=Kw/Ka1 For H2SO3 pKa1=1.81
This Ka is quite big (strong acid, very weak conjugate base) Practically you don't need to take it into account (pH will be very close to 7). If you truly need to find an exact value then you also have to consider the self-dissociation of water since it will be in the same order of magnitude. If you want that you can contact me.

For the second equivalence point
H2SO3 +2NaOH -> Na2SO3 + 2H2O

Now the stoichiometry is 1:2 for the acid:base
so 2 mole acid =1mole base=>
2*Ca*Va =Cb*Vb =>...
For the salt it is still 1:1 with respect to the acid, so you calculate as in (a)
The equilibrium is
SO3(-2) + H2O <=> HSO3(-) +OH(-)
with Kb=Kw/Ka2 and pKa2=6.97
Do the calculations as in (a)