Determine the values of Ka and Kb for the acid base dissociation reactions and report the...

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Chemistry

Determine the values of K_a and K_b for theacid base dissociation reactions and report the percent ionizationfor NH₃ and NH₄⁺. (this willrequire you to write the correctdissociation reaction for eachspecies in water, create ice diagram and write the correct Kexperession for each). If you can help, that will help alot!!!!
0.1M NH₄Cl: 6.53 pH, 0.1M NH₄OH: 10.28pH
0.01M NH₄Cl: 9.67 pH, 0.01M NH₄OH: 8.16pH

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3.6 Ratings (658 Votes)

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

NH4⁺ +H2Oâ†”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+H2Oâ†”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⁺ +H2Oâ†”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

Similar calculations for kb

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