Initially 2.0 moles of N_2 (g) and 4.0 moles of H_2 (g) were added to a 1.0-lite
ID: 508054 • Letter: I
Question
Initially 2.0 moles of N_2 (g) and 4.0 moles of H_2 (g) were added to a 1.0-liter container and the following reaction then occurred: 3 H_2 (g) + N_2 (g) 2NH_3 (g) The equilibrium concentration of NH_3 (g) = 0.55 moles/liter at 700. degree C. The value for K at 700. degree C for the formation of ammonia is: The following question refers to a 2.0-liter buffered solution created from 0.72 M NH_3 (K_b = 1.8 times 10^-5) and 0.26 M NH_4 F. What is the pH of this solution? Calculate the pH of a 0.47 M NH_3 (K_b = 1.8 times 10^-5) solution. In a solution prepared by dissolving 0.100 mole of propanoic acid in enough water to make 1.00 L of solution, the pH is observed to be 2.832. The K_a for propanoic acid (HC_3 H_5 O_2) is: At 500.0 K, one mole of gaseous ONCl is placed in a one-liter container. At equilibrium it is 5.3% dissociated according to the equation shown here: 2ONCl 2NO + Cl_2. Determine the equilibrium constant.Explanation / Answer
Ans 2 -
pOH = Kb + log [NH4F / NH3]
pOH = log-(1.8 x10-5) + log [0.26 / 0.72]
pOH = 4.745 + (-0.442)
pOH = 4.30
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 4.30
pH = -9.69
Ans 3-
NH3 + H2O <--------> NH4+ + OH-
C(1-X) CX CX
Kb = CX CX /C (1-X)
1.8 X10-5 = CX2 (1-X is very small)
1.8 X10-5 = 0.47X2
3.82 X 10-5 = X2
6.18 X 10-3 = X
OH- = CX = .47 x 6.18 X 10-3
OH- = 2.9 X10-3
pOH = -log OH-
pOH = 2.537
pH + pOH = 14
pH = 14 - 2.537
pH = 11.462