A is a solution of H3PO4 with pH of 1.46. — Analytical Chemistry Chemistry Question
Phosphoric acid
A is a solution of H3PO4 with pH of 1.46.
Calculate the molar concentrations of all species in solution A. Given that Ka values for H3PO4 are 7.2 ⋅ 10−3; 6.3 ⋅ 10−8 and 4.2 ⋅ 10−13, respectively.
Model Answer
H+ is used instead of H3O+ for clarity. The activities of the ions are ignored. [H+] is abbreviated as h in all calculations and acid constants for H3PO4 are written as K1, K2 and K3. As K1 >> K2 >> K3, only first dissociation step is considered.
H3PO4 ⇌ H+ + H2PO4-
As [H+][H2PO4-] / [H3PO4] = h^2 / (co - h) = (10^-1.46)^2 / (co - 10^-1.46) = K1 = 10^-2.14
Solving for co gives co = 0.200
The concentrations of the forms:
[H3PO4] = (h^3 * co) / (h^3 + h^2 K1 + h K1 K2 + K1 K2 K3) = 0.1653 mol dm-3
[H2PO4-] = (h^2 K1 co) / (h^3 + h^2 K1 + h K1 K2 + K1 K2 K3) = 0.0346 mol dm-3
[HPO42-] = (h K1 K2 co) / (h^3 + h^2 K1 + h K1 K2 + K1 K2 K3) = 6.29×10-8 mol dm-3
[PO43-] = (K1 K2 K3 co) / (h^3 + h^2 K1 + h K1 K2 + K1 K2 K3) = 7.56×10-19 mol dm-3
Mixing of 50 cm3 of solution A and 50 cm3 of NH3 solution (c = 0.4 mol dm-3) results in 100 cm3 of solution B. Calculate pH of solution B (pKa NH4+ = 9.24).
Model Answer
We have:
n(H3PO4) = 0.2 × 0.050 = 0.010 mol
n(NH3) = 0.4 × 0.050 = 0.020 mol
Hence the following reaction occurs: H3PO4 + 2 NH3 → (NH4)2HPO4
And [(NH4)2HPO4] = 0.010 / 0.100 = 0.1 mol dm-3
In solution B: (NH4)2HPO4 → 2 NH4+ + HPO42–
0.2 M 0.1 M
We have the following equilibria:
NH4+ ⇌ NH3 + H+
HPO42– + H+ ⇌ H2PO4–
H2PO4– + H+ ⇌ H3PO4
HPO42– ⇌ H+ + PO43–
A conservation of protons requires:
[H+] + 2 [H3PO4] + [H2PO4–] = [OH–] + [PO43–] + [NH3] (1)
In which [NH3] + [NH4+] = 0.2
[H3PO4] + [H2PO4–] + [HPO42–] + [PO43–] = 0.1
As pH of the solution is of about 7 – 9 so we can ignore the [H+], [OH–], [H3PO4] and [PO43–] in the equation (1):
[H2PO4–] = [NH3]
(h * 0.1) / (h + K2) = (KNH4 * 0.2) / (h + KNH4)
Solving for h gives h = 8.81 × 10-9 and pH = 8.06.
100 cm3 of solution B is mixed with 100 cm3 of Mg(NO3)2 solution (c = 0.2 mol dm-3). Determine if precipitate of NH4MgPO4 forms. The hydrolysis of Mg2+ is ignored and precipitation of NH4MgPO4 is assumed to be the only reaction. Ks(NH4MgPO4) = 2.5 ⋅ 10−13.
Model Answer
Mixing of B and Mg(NO3)2 solution leads to precipitation reaction:
NH4+(aq) + Mg2+(aq) + PO43– (aq) → NH4MgPO4(s)
[Mg2+] = 0.2 / 2 = 0.1 mol dm-3
As B is a buffer solution when it is diluted to twice the original volume, pH is virtually unchanged and is 8.06.
[NH4+] = (10^-8.06 * 0.1) / (10^-8.06 + 10^-9.24) = 0.094 mol dm-3
[PO43–] = (K1 K2 K3 * 0.05) / (h^3 + h^2 K1 + h K1 K2 + K1 K2 K3) = 2.06 * 10^-6 mol dm-3
The solubility product:
[NH4+][Mg2+][PO43–] = 0.1 × 0.094 × 2.06 ⋅ 10^-6 = 1.93 ⋅ 10^-8 > 2.5 ⋅ 10^-13
Therefore, the precipitation occurs.
Calculate the solubility (mol·dm−3) of Ca3(PO4)2 if Ks(Ca3(PO4)2) = 2.22 ⋅ 10−25. (Hint: The hydrolysis of Ca2+ is ignored).
Model Answer
We have: Ca3(PO4)2 ⇌ 3 Ca2+ + 2 PO43–
Assume that the hydrolysis of PO43– can be ignored, the solubility so of Ca3(PO4)2 can be calculated as follows:
Ks = [Ca2+]^3[PO43–]^2 = (3 so)^3(2 so)^2 = 108 so^5 = 2.22 ⋅ 10−25
Solving for so gives so = 4.6 ⋅ 10−6 mol dm-3
However, the hydrolysis of PO43– cannot be ignored due to its rather strong basicity (pKb = 14 – pKa = 14 – 12.38 = 1.62)
PO43– + H2O ⇌ HPO42– + OH- (1)
We can ignore the hydrolysis of HPO42– (pKb = 14 – 7.20 = 6.80) and H2PO4– (pKb = 14 – 2.14 = 11.86).
According to (1): [HPO42–] + [PO43–] = 2 s (2)
As [PO43–] is very small, it can be ignored in (2). It can alternatively be calculated as follows:
Let x is the concentrations of HPO42– and OH–, [HPO42–] = [OH–] = x
We have: x^2 / (2 × 4.6 × 10^-6 - x) = 10^-1.62 = 0.024
Solving for x gives x = 9.19 ⋅ 10^-6 → [PO43–] = 0.01 ⋅ 10−6 mol dm-3
Therefore we can assume that [HPO42–] = [OH–] = 2s and [PO43–] is determined based on K3:
[PO43-] = (K3 * [HPO42-]) / [H+] = (K3 * [HPO42-] * [OH-]) / Kw = 167s
The solubility s of Ca3(PO4)2: Ks = 2.22 ⋅ 10^-25 = (3 s)^3 (2 × 167 s^2)^2 = 3012012 s^7
⇒ s = 3.6 ⋅ 10^-5
We can see that solubility of Ca3(PO4)2 increases about 10 times due to the hydrolysis of PO43–.