Question

5) Phosphoric acid (H3PO4) is added to deionized water at a total concentration of 0.01 M and then sodium hydroxide is added until the pH is increased to 5.0. a) What is the value of the first ionization fraction (a for H2PO4)? b) What molar concentration of H2PO4' would you expect at this pH?

Answer 1

5.Ans :-

(a). The ICE table of phosphoric acid is :

........................................H₃PO₄ (aq) <--------------------------------> H₂PO₄^- (aq) ................+...................H⁺ (aq)

Initial ................................0.01 M......................................................0.0 M..............................................0.0 M

Change ......................-0.01α₁ .........................................................+0.01α₁.........................................+0.01α₁

Equilibrium....................0.01(1-α₁​​​​​​​) M..................................................0.01α₁​​​​​​​............................................0.01α₁​​​​​​​

Where α₁​​​​​​​ = first ionization fraction

At equilibrium, [H⁺] = 10^-pH = 10^-5 M (because, pH = - log [H⁺]

Therefore, 0.01α₁​​​​​​​ = 10^-5

and

α₁​​​​​​​ = 10^-5/0.01

α₁​​​​​​​ = 10^-3

Therefore, first ionization fraction (α₁) = 0.001

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​​​​​​​(b).

Now, Equilibrium concentration of H₂PO₄^- = 0.01α₁ = 0.01 x 0.001 = 10^-5

Therefore,

Required Molar concentration of H₂PO₄^- = 10^-5 M