Answer: approximately 2.8
Step-by-step explanation:
To determine the pH you need to find the concentration of H₃O⁺ after the reaction is completed.
This is how much of the acid was in excess and did not react.
Therefore, first we will calculate the amount of excess H₃PO₄, next will use the equilibrium constant of the acid to calculate the amount of H₃O⁺, which will give us the pH of the solution:
1) Chemical equation:
3KOH + H₃PO₄ → K₃PO₄ +3H₂O
2) Mole ratio:
3 mol KOH : 1 mol H₃PO₄
3) Convert the given data into number of moles
a) KOH:
• Volume: 3V
• Molarity: 7mM (this is 7 milimolar)
• Number of moles: M = n / V ⇒ n = M×V = 3V×7mM = 21V milimoles
b) H₃PO₄:
• Volume: 1V
• Molarity: 15 mM
• Number of moles: M = 1V × 15 mM = 15 milimoles
4) Limiting reactant: find the moles that react using the theoretical mole ratio:
3 mol KOH / 1 mol H₃PO₄ = 21V milmoles KOH / x ⇒
x = 21V × 1 / 3 milimoles H₃PO₄ = 7V milimoles H₃PO₄
Since there are 15V milimoles of H₃PO₄, only 7V reacted (limiting reactant) and 15Vmilimoles - 7Vmilimoles = 8Vmilimoles are left over.
5) Calculate the molar concentration of the 8V milimoles of of acid in excess:
- M = number of moles / volume of the solution in liters
- Volume of the solution: 3V + 1V = 4V
- M = 8Vmilimoles / (4V) = 2 milimoles = 0.002 moles
6) Now, work with the equilibrium equation for phosphoric acid, under these assumptions:
- Phosphoric acid is a medium strong acid and only the first ionization is quantitatively important to impact the pH:
H₃PO₄ + H₂O → H₃O+ + H₂PO₄⁻
- First Ka is 7.1 × 10⁻³ = 0.0071
H₃PO₄ + H₂O ⇄ H₃O+ + H₂PO₄⁻
0.002 – x x x
- Ka = x² / (0.002 – x) = 0.0071
⇒ x² = 0.0071×0.002 – 0.0071x
⇒ x² + 0.0071x – 0.0000142 = 0
Using quadratic formula: x = 0.00162711
[H3O+] = 0.00162711
pH = - log [H3O+] = - log (0.00162711) = 2.8 ← answer