If 3 volumes of 7 mm koh are mixed with 1 volume of 15 mm phosphoric acid (h3po4), what will be the ph of the final mixture

approximately 2.8

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:

6) Now, work with the equilibrium equation for phosphoric acid, under these assumptions:

         H₃PO₄ + H₂O  → H₃O+ + H₂PO₄⁻

         H₃PO₄           + H₂O  ⇄    H₃O+ + H₂PO₄⁻

        0.002 – x                             x            x

        ⇒ 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

the answer is a and have a great day