Question

Citric acid is triprotic (H3Cit) with acidity constants (at 25 °C) of pKa1 = 3.13, pKa2 = 4.76, and pKa3 = 6.40.

a) Write out the expressions for α_0, α_1, α_2, and α_3 as a function of Ka values and {H+}.

b) Identify the pH range in which each citrate species (H3Cit, H2Cit–, HCit2–, and Cit3–) predominates.

c) A solution is prepared by dissolving 0.10 mol of Na2HCit in enough deionized water to give a total volume of 1.00 L. Promptly after dissolving in water, the salt Na2HCit dissociates “completely” according to the following reaction: Na2HCit ⇌ 2Na+ + HCit– What is the equilibrium pH of this solution? Calculate the pH manually while assuming ideal behavior.

Answer 1

So we have a tripotic citric acid(H₃Cit ), We have to write its ionic reaction and reach to its details.We have already been provided with the values of pK_a1 , pK_a2 and pK_a3 .

H₃Cit  $\rightarrow$ H⁺ + H₂Cit^-

1-$\alpha$₀$\alpha$₀   $\alpha$₀ K_a1 = [ H₂Cit^- ] [H⁺ ] / [H₃Cit ] $\Rightarrow$$\alpha$₀ [H⁺ ] / 1-$\alpha$₀ =  K_a1

$\Rightarrow$$\alpha$₀ / 1-$\alpha$₀ =   K_a1 /  [H⁺ ] $\Rightarrow$$\alpha$₀ =  K_a1 /  K_a1 +  [H⁺ ]

Similarly, H₂Cit^-$\rightarrow$ HCit^2- + H⁺ ,  $\Rightarrow$ K_a2 =   [ HCit^2- ] [H⁺ ] / [H₂Cit^- ]

$\alpha$₀ - $\alpha$₁   $\alpha$₁   $\alpha$₁ +  $\alpha$₀$\Rightarrow$   [ $\alpha$₁ +  $\alpha$₀ ] [$\alpha$₁ ] / [ $\alpha$₀ - $\alpha$₁ ] =  K_a2 substituting value of  $\alpha$₀ from the above expression, we get  $\alpha$₁ = ( K_a2 /  K_a2 + H⁺ ) ( K_a1 /  K_a1 +  [H⁺ ] )

Again, HCit^2-$\rightarrow$ Cit^3- + H⁺  $\Rightarrow$  K_a3 =  [ Cit^3- ] [H⁺ ] / [HCit^2- ]  

$\alpha$₁ - $\alpha$₂$\alpha$₂     $\alpha$₂ + $\alpha$₁ +  $\alpha$₀$\Rightarrow$ K_a3 = ($\alpha$₂ + $\alpha$₁ +  $\alpha$₀ ) ( $\alpha$₂ ) / ( $\alpha$₁ - $\alpha$₂ ) , Substituting the values of $\alpha$₁ &  $\alpha$₂ from the above expressions, we get :

$\alpha$₂ = (K_a3 / K_a3 +  H⁺ )( K_a2 /  K_a2 + H⁺ ) ( K_a1 /  K_a1 +  [H⁺ ] )

c) Now we have a solution of 0.1 mole of Na₂HCit in 1 litre , so it implies a concentration of 0.1 M, Now writing the ionic equation, we get

H₂Cit^-  $\rightarrow$ HCit^2- + H⁺

0.1-x x x $\Rightarrow$ x² / 0.1-x = K_a1 , Now, ignoring x with respect to 0.1, we get

x² / 0.1 = K_a1 , x =  (K_a1 / 0.1)^0.5 , pH = -logx = 0.5 ( -log K_a1 + log 0.1 ) = 0.5 ( pK_a2 - 1) = 0.5 (4.76-1) = 1.88