Answer:
Na⁺, A³⁻
Explanation:
For easy calculation, let's say you have mixed 1 L of each of these solutions to make 4 L total.
Citric acid (H₃A) has three acidic hydrogens.
The equilibria are
1. H₃A + H₂O ⇌ H₃O⁺ + H₂A⁻; pKₐ = 3.08
2. H₂A⁻ + H₂O ⇌ H₃O⁺ + HA²⁻; pKₐ = 4.44
3. HA²⁻ + H₂O ⇌ H₃O⁺ + A³⁻; pKₐ = 5.40
4. A³⁻ + H₂O ⇌ OH⁻ + HA²⁻ ; pKb = 8.60
5. 2H₂O ⇌ H₃O⁺ + OH⁻; pKw = 14.00
Now, we adjust to pH 9.
Let's look at the weakest acid (Equation3),
pH = pKa + log([A³⁻]/[HA²⁻])
9 = 5.40 + log([A³⁻]/[HA²⁻])
log([A³⁻]/[HA²⁻])= 3.6
[A³⁻]/[HA²⁻] = 10^3.6 = 4000
[A³⁻]= 4000[HA²⁻]
In other words, at pH 9, the weakest acid is completely neutralized. Then, the stronger acids are also completely neutralized.
It takes 3 mol of NaOH to neutralize the H₃A, 2 mol of NaOH for the NaH₂A, and 1 mol NaOH for the Na₂HA.
So, 6 mol of NaOH will neutralize the three acids, bring the pH to 9, and make a total volume of 10 L.
The final solution contains the species: H₃A, H₂A⁻, HA²⁻, A³⁻, Na⁺, H₃O⁺, and OH⁻.
Now we must assess their relative amounts.
Na⁺: 6 mol for neutralization + 6 mol in the original solution = 12 mol.
[Na⁺] = 1.2 mol·L⁻¹
A³⁻: 1 mol from each of the four solutions = 4 mol A³⁻.
[A³⁻] = 0.4 mol·L⁻¹
The other species all have concentrations less than 10⁻⁴ mol·L⁻¹.
The major species are Na⁺ and A³⁻.
