NH3 (aq) + H2O (l) <--> NH4 (aq) + OH (aq)
A 0.0124 M ammonia solution is introduced into the above equilibrium reaction is 25 degrees C. When a system achieves equilibrium, the hydroxide concentration is measured to be 4.64x10^4 M. Calculate the equilibrium constant of this system. So I did an Ice box and ignored H2O since is liquid and is not supposed to be included in K. However I end up with 0 = X^2 + X -0.0124 and would have to do the quadratic equation, however I have a feeling that this is wrong since I just starting doing this. Is NH3 's Equilibrium I get .0124 - x. and for NH4 and OH's I get 4.64X10^4 since they have equal mol ratios. Please help I need to get the answer
A 0.0124 M ammonia solution is introduced into the above equilibrium reaction is 25 degrees C. When a system achieves equilibrium, the hydroxide concentration is measured to be 4.64x10^4 M. Calculate the equilibrium constant of this system. So I did an Ice box and ignored H2O since is liquid and is not supposed to be included in K. However I end up with 0 = X^2 + X -0.0124 and would have to do the quadratic equation, however I have a feeling that this is wrong since I just starting doing this. Is NH3 's Equilibrium I get .0124 - x. and for NH4 and OH's I get 4.64X10^4 since they have equal mol ratios. Please help I need to get the answer
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"NH3 's Equilibrium I get 0.0124 - x. and for NH4 and OH's I get 4.64X10^-4 "
This statement is right but the problem is simpler than you're making it. I'm going to assume you mean 4.64x10^-4 or 0.000464M since you can't have more reactants than products. Note that x= 0.000464M. Since the reaction constituents are all in equal molar ratios the amount of NH3 reacted = the amount of each product formed.
So at equil. NH3= 0.0124-0.000464=0.0119 M
K=[NH4+][OH-]/[NH3]= [0.000464][0.000464]/[0.0119]=1.80 x 10^-5
This statement is right but the problem is simpler than you're making it. I'm going to assume you mean 4.64x10^-4 or 0.000464M since you can't have more reactants than products. Note that x= 0.000464M. Since the reaction constituents are all in equal molar ratios the amount of NH3 reacted = the amount of each product formed.
So at equil. NH3= 0.0124-0.000464=0.0119 M
K=[NH4+][OH-]/[NH3]= [0.000464][0.000464]/[0.0119]=1.80 x 10^-5