1) Kb for NH3 is 1.8x10^-5. What is the pOH of a .20M aqueous solution of NH4Cl
ID: 954950 • Letter: 1
Question
1) Kb for NH3 is 1.8x10^-5. What is the pOH of a .20M aqueous solution of NH4Cl at 25 degrees Celsius? (Please show work)2) Determine the pH of a 0.15 M aqueous solution of KF. For hydrofluoric acid, Ka=7.0x10^-4. (Please show work)
3) An aqueous solution of _____ will produce a neutral solution.
A) NaNO2 B) LiNO3 C) KF D) Rb2CO3 E) NH4NO3
1) Kb for NH3 is 1.8x10^-5. What is the pOH of a .20M aqueous solution of NH4Cl at 25 degrees Celsius? (Please show work)
2) Determine the pH of a 0.15 M aqueous solution of KF. For hydrofluoric acid, Ka=7.0x10^-4. (Please show work)
3) An aqueous solution of _____ will produce a neutral solution.
A) NaNO2 B) LiNO3 C) KF D) Rb2CO3 E) NH4NO3
2) Determine the pH of a 0.15 M aqueous solution of KF. For hydrofluoric acid, Ka=7.0x10^-4. (Please show work)
3) An aqueous solution of _____ will produce a neutral solution.
A) NaNO2 B) LiNO3 C) KF D) Rb2CO3 E) NH4NO3
Explanation / Answer
1) Ka of Nh4+ = Kw/kb = 10^-14 /1.8x10^-5 = 5.55 x 10^-10
NH4+ (aq) <---> Nh3 (aq) + H+ (aq) ,
at equilibrium [NH4+]= 0.2-X , [NH3] = [H+] = X
Ka = [Nh3] [H+] /[NH4+]
5.55 x 10^-10 = ( X) ( X) / ( 0.2-X)
X= [H+] =1.054 x 10^-5 , pH = -log [H+] = -log ( 1.054x10^-5) = 5
pOH = 14-pH = 14-5 = 9
2) F- (aq) + H2O (l) <--> HF (aq) + OH- (aq) , Kb = Kw/KaHf = 10^-14 / 7x10^-4 = 1.4286 x 10^-11
at equilibrium [F-] = 0.15-X , [HF] =[OH-] = X
Kb = [HF][OH-] /[F-]
1.4286 x 10^-11 = ( X) ( X) / ( 0.15-X)
X = [OH-] = 1.464 x 10^-6 , pOH = -log ( 1.464 x 10^-6) = 5.835
pH = 14-5.835= 8.165
3) storng acid and strong base combination gives salt which is neutral , hence its LiNO3 , since LioH is strong base and HNO3 is strong acid
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