Question

(19) Ammonium formate, NH4(HCOO), dissolves in water to give ammonium ions and formate ions. These ions then undergo an unfavorable and endothermic Brønsted acid- base reaction according to this equation:

NH4+ (aq) + HCOO– (aq) NH3 (aq) + HCOOH (aq)

Keq < 1 (at 298 K)

∆H_orxn > 0

(a) (8 pts.) Calculate Keq and ∆G_orxn (at 298 K) for equation (19).

(b) (6 pts.) Qualitatively, would one expect ∆S_orxn for equation (19) to be positive or negative? Briefly justify your answer.

(c) (10 pts.) If 1.00 mol NH4(HCOO) is dissolved in 1.00 L water at 298 K, what is the pH of the solution?

(d) (7 pts.) The solution from part (c) (1.00 mol NH4(HCOO) in 1.00 L water) can be made into an effective buffer at pH = 3.00 by the addition of either NaOH or HCl. Specify whether NaOH or HCl would be required, and whether more than 0.500 or less than 0.500 mol of the reagent would be needed.

(e) (8 pts.) Consider the solution from part (c) (1.00 mol NH4(HCOO) in 1.00 L water at 298 K). For each of the changes below, circle the effect on the concentration of NH3 (aq) at equilibrium after the change is made, and briefly give the reasoning for your choice.

NH4+ (aq) + HCOO– (aq) NH3 (aq) + HCOOH (aq)

Keq < 1 (at 298 K)

∆H_orxn > 0
(i) The temperature of the solution is increased to 340 K.

[NH3] increases [NH3] decreases [NH3] does not change

Explanation:

(ii) The volume of the solution is increased to 2.00 L by adding 1.00 L water.

[NH3] increases [NH3] decreases [NH3] does not change

Explanation:

Answer 1

a) The given equation is

NH₄⁺ (aq) + HCOO^- (aq) <======> NH₃ (aq) + HCOOH (aq)

K_eq < 1 (at 298 K) and ΔH⁰_rxn

Since exact values are not available, we can write down the expression for K_eq for the above equilibrium in terms of the molar concentrations of the reactants and products as

K_eq = [NH₃][HCOOH]/[NH₄⁺][HCOO^-]

Given K_eq < 1 , we can say that ln K_eq < 0, i.e., the natural logarithm of K_eq will be negative.

We know that ΔG⁰_rxn = -RTln K_eq. Since T = 298 K and R = 8.314 J/mol.K and ln K_eq < 0, therefore, ΔG⁰_rxn > 0 (two negatives make a positive) (ans).

b) We know that

ΔG⁰_rxn = ΔH⁰_rxn – T.ΔS⁰_rxn

Since both ΔG⁰_rxn, ΔH⁰_rxn > 0 and T > 0, therefore ΔS⁰_rxn must have a low positive value.

c) Write down the dissociation equation and ICE chart as

NH₄(HCOO) (aq) <=====> NH₄⁺ (aq) + HCOO^- (aq)

Initial                                     (1.00 mol/1 L) = 1.00 M          0                     0

Change                                           -x                                      +x                   +x

Equilibrium                               (1.00 – x)                                x                     x            

We have a weak acid, NH₄⁺, hence we can use the K_a for ammonium ion, which is 5.8*10^-10

Write down the expression for K_a as

K_a = [NH₄⁺][HCOO^-]/[NH₄HCOO] = (x).(x)/(1.00 – x)

===> 5.8*10^-10 = x²/(1 – x)

Since K_a is very small, of the order of 10^-10, we can assume x << 1.00 and write,

5.8*10^-10 = x²/1

====> x² = 5.8*10^-10

===> x = 2.408*10^-5

Therefore, [NH₄⁺] = 2.408*10^-5 M. Note that NH₄⁺ -------> NH₃ + H⁺ so that [H⁺] = 2.408*10^-5 M.

Therefore, pH = -log [H⁺] = -log (2.408*10^-5) = 4.618 ≈ 4.62 (ans).

d) The pH of the solution of ammonium formate is calculated to be 4.62 and we want to make a buffer with pH 3.00. Since we are to reduce the pH of the solution, we must add HCl (strong acid) which will bring down the pH. Let x mol of HCl be added and the total volume is 1 L so that the molar concentration of added H⁺ = x M.

Write down the neutralization reaction as

NH₄HCOO (aq) + HCl (aq) -----> NH₄Cl (aq) + HCOOH (aq)

Initial                                     1.00                      x                         0                      0

Change                                    -x                      -x                       +x                    +x

Equilibrium                      (1.00 – x)                  0                         x                      x

Use Henderson-Hasslebach equation as

pH = pK_a + log [NH₄HCOO]/[HCOOH]

====> 3.00 = 4.62 + log (1 – x)/(x)

===> -1.62 = log (1-x)/x

===> (1-x)/x = antilog (-1.62) = 0.02398

===> 1 – x = 0.02398x

===> 1 = 1.02398x

===> x = 1/1.02398 = 0.97658 ≈ 0.98

Therefore, 0.98 mole of HCl must be added to the solution (ans).