Question

1. Calculate the pH of the solution obtained by adding 0.10 moles of NaOH to a 1 liter solution of 0.5 M H2SO3. The Ka for H2SO3= 1.71X10^-2. (YOU MAY USE THE HENDERSON HASSELBACK EQUATION).

2. Consider a 1 Liter solution that is 0.25 mole in HF and 0.5 mole solution in NaF. Use the Henderson Hasselbalck equation . Ka for HF= 7.2X10^-4

a. Calculate the pH for the solution.

b. Calculate the pH of the solution upon addition 0.02 mole HNO3

c. Calculate the pH of the solution upon addition of 0.04 mole KOH.

3. A solution is 0.012 M in Pb(NO3)2 and 0.20 M in Sr(NO3)2 . Solid Na2SO4 is added until a precipitate of PbSO4 just begins to form

a. Calculate the concentration of sulfate ion at this point.

Ksp for PbSO4 is 1.8x10^-8 and for SrSO4 is 2.8x10^-7

Answer 1

1. Henderson equation is: pH = pKa + log [salt]/[acid]

The balanced equation is : 2NaOH + H₂SO₃ = Na₂SO₃ + 2H2O

According to the equation, 2 mol of NaOH gives 1 mol of salt(Na₂SO₃).

0.1 mol of NaOH gives 0.05 mol of salt.

Remaining amount of acid = 0.5 - 0.05 = 0.45 mol

As the solution is 1 lit, concentration of acid = 0.45M and concentration of salt = 0.05 M

pH = pKa + log [salt]/[acid]

= -log(1.71X10^-2) + log 0.05/0.45

=0.82

2. a. pH = pKa + log [salt]/[acid]

= -log(7.2X10^-4) + log 0.5/0.25

=3.44

b. H+ from HNO₃ will react with NaF to form HF, thus concentration of salt will decrease and that of the acid increase.

NaF + H⁺ = HF + Na⁺

[NaF] = 0.5-0.02 = 0.48

[HF] = 0.25-0.02

pH = pKa + log [salt]/[acid]

= 3.14 + log 0.48/0.27

=3.39

c. Similar as above. The reaction is: HF + OH^- = H₂O + F^-

[NaF] = 0.5+0.04=0.54

[HF] = 0.25-0.04 = 0.21

pH = pKa + log [salt]/[acid]

= 3.14 + log 0.54/0.21

=3.55

3. Ksp of lead sulfate is smaller than strontium sulfate. So lead sulfate will be first to precipitate.

At the point of saturation lead sulfate will be about to precipitate. At this point,

[SO₄^2-][Pb²⁺] = Ksp

So, [SO₄^2-]= Ksp / [Pb²⁺]

= 1.8 * 10^-8 / 0.012 = 1.5 * 10^-6