Answer:
a) pH = 9.8
b) 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]
c) decrease
Explanation:
The equilibriums involved in this question are:
C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + OH⁻ (1) Kb₁ =[HC₃H₂O₄⁻][OH⁻]/[C₃H₂O₄²⁻ ]
HC₃H₂O₄⁻ + H₂O ⇄ H₂C₃H₂O₄ +OH⁻ (2) Kb₂=[ H₂C₃H₂O₄]/[OH⁻]/[HC₃H₂O₄⁻]
The kas for malonic acid, H₂C₃H₂O₄, from reference tables are:
Ka (H₂C₃H₂O₄ )= 1.4 x 10⁻³
Ka ( HC₃H₂O₄⁻ ) = 2.0 x 10⁻⁶
a) We can calculate the Kbs for the conjugate bases of the weak malonic acid from Kw = Ka x Kb
Kb (C₃H₂O₄²⁻) = 10⁻¹⁴ / 2.0 x 10⁻⁶ =5.0 x 10⁻⁹
Kb (HC₃H₂O₄⁻)= 10⁻¹⁴ / 1.0 x 10⁻³ = 7.1 x 10⁻¹²
Given the magnitudes of the Kbs ( Kb₂ is approximately 1000 times Kb1 ) , to calculate pOh we can neglect the contribution from (2). We then treat this problem as any equilibrium:
[K₂C₃H₂O₄] = 25 g/180.2 g/mol / 0.455 L = 0.30 M
Conc C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + 0H⁻
I 0.30 0 0 0
C -x +x +x
E 0.30 - x x x
Kb (C₃H₂O₄²⁻) = [ HC₃H₂O₄⁻ ] [OH⁻]/ [ C₃H₂O₄²⁻] = x² / 0.30 - x ≅ x² /0.30
x² /.030 = 5.0 x 10⁻⁹ ⇒ x = √(0.30 x 5.0 x 10⁻⁹ ) = 7.1 x 10⁻⁵ = [ÒH⁻]
(Verifying our approximation was good 7.1 x 10⁻⁵ / 0.30 = 2.4 x 10⁻⁴ so our approximation checks)
pOH = -log 7.1 x 10⁻⁵ = 4.2
pH = 14 -4.2 = 9.8
b) To answer this part we take equilibrium (2 ) and set up our usual ICE table to solve for the concentration of malonic acid:
Conc (M) HC₃H₂O₄⁻ + H₂O ⇄ H₂C₃H₂O₄ + OH⁻ (2)
I 7.1 x 10⁻⁵ 0 0
C -x +x +x
E 7.1 x 10⁻⁵ -x x x
7.1 x 10⁻⁵ - x ≅ 7.1 x 10⁻⁵
[ H₂C₃H₂O₄ ] [OH⁻] / [HC₃H₂O₄⁻] = Kb₂ = 7.1 x 10⁻¹² = x² / 7.1 x 10⁻⁵
x = √(7.1 x 10⁻¹² x 7.1 x 10⁻⁵) = 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]Again our approximation checks since [HC₃H₂O₄⁻] is almost 1000 times [ H₂C₃H₂O₄ ]
c) From eqn (1) :
C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + OH⁻
The salt K₂C₃H₂O₄ will react completely with the added acid, thereby decreasing the C₃H₂O₄²⁻ concentration, and according to Le Chateliers principle the system will shift to the left and the OH⁻ at equilibrium will decrease ( as also does [HC₃H₂O₄⁻] ) therefore the pOH will increase and the pH will decrease ( less OH⁻ higher pOH, smaller pH )
