Question

Determine the values of K_a and K_b for the acid base dissociation reactions and report the percent ionization for NH₃ and NH₄⁺. (this will require you to write the correctdissociation reaction for each species in water, create ice diagram and write the correct K experession for each). If you can help, that will help alot!!!!

0.1M NH₄Cl: 6.53 pH, 0.1M NH₄OH: 10.28 pH

0.01M NH₄Cl: 9.67 pH, 0.01M NH₄OH: 8.16 pH

Answer 1

1)

0.1M NH4Cl: 6.53 pH, 0.1M NH4OH: 10.28 pH

NH4⁺ +H2O↔NH3+H3O+

pH=6.53=-log[H3O+]

[H3O+]=10^-pH=10^-6.53=2.95*10^-7M

Ka=[H3O+][NH3]/[ NH4⁺]

	[NH4+]	[NH3]	[H3O+]
initial	0.1M	0	0
change	-x	+x	+x
equilibrium	0.1-x	x	x

Ka=[H3O+][NH3]/[ NH4⁺]=(x)(x)/(0.1-x)

Or, ka=(x)(x)/(0.1-x) [ignore x<< than 0.1)

Ka=x*x/0.1

But x=(2.95*10^-7M=[H3O+]

Ka=(2.95*10^-7)^2/0.1=87.025*10^-14=8.7*10^-13

Ka=8.7*10^-13

% ionization of NH4+=2.95*10^-7/0.1*100=2.95*10^-6 %

Solve for kb

NH3+H2O↔NH4+ +OH-

Kb=[NH4+][OH-]/[NH3]

Given [NH3]=0.1M

pH=10.28

pOH=14-10.28=3.72

[OH-]=10^-pH=10^-3.72=1.9*10^-4M

	[NH3]	[NH4+]	[OH-]
initial	0.1M	0	0
change	-x	+x	+x
equilibrium	0.1-x	x	x

Kb=x*x/0.1-x

Or,kb=x^2/0.1

x= 1.9*10^-4M=[OH-]

kb=(1.9*10^-4)^2/0.1=36.1*10^-4=3.61*10^-3

kb=3.61*10^-3

% ionization=x/0.1*100=1.9*10^-4M/0.1*100=1.9*10-3=0.0019%

2) 0.01M NH₄Cl: 9.67 pH, 0.01M NH₄OH: 8.16 pH

NH4⁺ +H2O↔NH3+H3O+

pH=9.67=-log[H3O+]

[H3O+]=10^-pH=10^-9.67=2.14*10^-10M

Ka=[H3O+][NH3]/[ NH4⁺]

	[NH4+]	[NH3]	[H3O+]
initial	0.01M	0	0
change	-x	+x	+x
equilibrium	0.01-x	x	x

Ka=[H3O+][NH3]/[ NH4⁺]=(x)(x)/(0.01-x)

Or, ka=(x)(x)/(0.01-x) [ignore x<< than 0.01)

Ka=x*x/0.1

But x=2.14*10^-10M=[H3O+]

Ka=(2.14*10^-10M)^2/0.1==4.58*10^-19

Ka=4.58*10^-19

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