Asolution was made by dissolving 25 g of potassium malonate k2c3h2o4 (mm=180.2 g/mol) in water. the total volume of that solution is 455 ml.

a) calculate the ph of that solution

b) calculate the concentration of malonic acid (h2c3h2o4) in that solution

c) what would happen to concentration of malonic acid (increase or decrease in the above solution if ph is decreased by addition of 12 m hcl?

explain your answer using net ionic equations and le chatelier's principle.

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 )