(3) a 10.00-ml sample of 0.1000 m kh2po4 was titrated with 0.1000 m hcl ka for phosphoric acid (h3po4): ka1= 7.50x10-3; ka2=6.20x10-8; ka3= 4.20x10-10 calculate the ph after addition of 10.00 ml of hcl to the kh2po4 solution?

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!

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