Common Units, Constants, and Math Formulas. R = 8.31 J/mol K (Pa m^3/mol K) = 8.
ID: 1066004 • Letter: C
Question
Common Units, Constants, and Math Formulas. R = 8.31 J/mol K (Pa m^3/mol K) = 8.31 kJ/mol K kPa m^3/mol K) R = 0.00831 kJ/mol K (kPa m^3/mol K) R = 8.31 times 10^-5 bar m^3/mol K 1.00 atm = 1.01 bar = 101 kPa = 760 mm Hg = 14.7 psi d/dx(f/g_ = g(df f fx = f(dg/dx)/g^2 d/dx(fg) = f dg/dx + g df/dx d(x^0/dx = ax^0-1 A liquid water stream contains 0.100 mol/L of NH_4CI and nothing else. The NH_4CI is completely dissociated. The stream is exposed to air containing 1000 ppm NH_3 at 25 degree C, and 2.0 atm and brought to equilibrium. Using the following data at 25 degree C. calculate the concentrations (mol/L) of NH_3. NH_4+, OH-, and H+ in the liquid phase at equilibrium. The effects of CO_2 in the air may be neglected. k_H = [NH_3]/y_NH_3 = 10^1/76 mol/liter-atm K = [NH_3][H+]/NH^+_4 = 10^-9.26 mol/liter K_w = [H+][OH-] = 10^-14 Mol^2/liter^2Explanation / Answer
1)Henry law constant=KH=10^1.76 mol/L atm=[NH3]/yNH3*P whereP= total pressure=2.0atm ,yNH3=1000*10^-6=10^-3
pNH3=partial pressure of NH3=yNH3*P
[NH3]=(10^1.76 mol/L atm)*10^-3 *2atm=115.088*10^-3=0.115 mol/L [solubility of NH3 in water]
2)Now in water the equilibrium exists as-
NH4+ <------>NH3 +H+
ICE table
[NH4+] [NH3] [H+]
Initial 0.1 mol/L 0.115mol/L 0
change -x +x +x
equilibrium 0.1-x 0.115+ x x
K=[NH3][H+]/[NH4+]=10^-0.20M=(0.115+x)*x/(0.1-x) [x<<<than 0.1 as K is very small so very less dissociation takes place]
K=0.115*x/0.1=10^-9.26
x=4.77 *10^-10 M=[H+]eq
[NH3]=0.115M+x=0.115=1.15*10^-1 mol/L
[OH-]=kw/[H+]=10^-14/4.77 *10^-10=0.209*10^-4=2.096*10^-5 M
pH=-log [H+]=-log (4.77 *10^-10)=9.32
[NH4+]=0.1M=1*10^-1 mol/L
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.