Answer:
The pH of this solution is 1,350
Explanation:
The phosphoric acid (H₃PO₄) has three acid dissociation constants:
HPO₄²⁻ ⇄ PO4³⁻ + H⁺ Kₐ₃ = 4,20x10⁻¹⁰ (1)
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸ (2)
H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)
The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).
We need to know what is the concentration of H⁺ for calculate the pH.
The moles of H₂PO₄⁻ are:
10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol
The moles of H⁺ are, in the same way:
10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol
So:
H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)
X mol ⇄ (1x10⁻³-X) mol + (1x10⁻³-X) mol (4)
The chemical equilibrium equation is:
Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]
So:
7,50x10⁻³ = (1x10⁻³-X)² / X
Solving the equation you will obtain:
X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0
Solving the quadratic formula you obtain two roots:
X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles
X = 1,065x10⁻⁴
So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol
The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)
Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M
pH is -log [H⁺]. So the obtained pH is 1,350
I hope it helps!
is = 47 g
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