A 189.7mL sample of 0.375M acetic acid (CH3CO2H) was allowed to react with 2.085L of gaseous ammonia at 25 ∘C & a pressure of 634.2 mmHg .?
a)Assuming no change in the volume of the solution, calculate the pH. Values of equilibrium constants are listed below.
Acetic Acid Ka = 1.8 x 10^-5
NH3 Ka = 1.8 x 10^-5
b)Assuming no change in the volume of the solution, the equilibrium concentrations of all species present (CH3CO2H, CH3CO−2, NH3, NH+4, H3O+, and OH− ).
More Questions From This User See All
Answers & Comments
Verified answer
a) First, calculate moles of each reactant:
moles HAc = 0.1897 L X 0.375 mol/L = 0.0711 mol HAc
Use the ideal gas law to calculate moles NH3:
P = 634.2 mm Hg/ 760 mm Hg/atm = 0.8345 atm
PV = n RT
(0.8345 atm) (2.085 L) = n (0.08206 Latm/molK) (298.15 K)
n = 0.0711 mol
So, these are reacting in a 1:1 ratio, and the final solution will contain 0.0711 mol of ammonium acetate in 0.375 L, so [NH4+] = [CH3COO-] = 0.190 M
Now, there are two ionizations going on in the solution:
NH4+ <--> H+ + NH3
CH3COO- + H2O <--> CH3COOH + OH-
(NOTE: The values that you provided in your problem are the values for Ka of acetic acid and the Kb of ammonia.)
Ka of NH4+ = 1X10^-14 / Kb = 5.6X10^-10
Kb of CH3COO- = 1X10^-14 / Ka = 5.6X10^-10
Now, since these two ionization constants have the same magnitute, [H+] from the first will equal [OH-] from the second. This defines a neutral solution, so pH = 7.0.
b) Since you know that [H+] = [OH-] = 1X10^-7, you can just plug those into the expressions for the appropriate Ka or Kb and solve for [NH3] and [CH3COOH]. The concentrations of NH4+ and CH3COO- will effectively be 0.190 M