A sample of 0.25M NH, is placed into a sealed container and allowed to come to e

Question

A sample of 0.25M NH, is placed into a sealed container and allowed to come to equilibrium. Calculate the final concentrations of all the species in the system. (Hint: You may want to employ the "x is small approximation in this problem and don't forget to check it.) d. 2. Methanol can by decompose to carbon monoxide and hydrogen CH:0H(g) CO()+2H2(8) K 0.232 for this reaction at 250°C. A chamber at 250°C was filled with 0.350M methanol and allowed to come to equilibrium. Find the equilibrium concentrations of all three species. Note that you will get a cubic equation. Use a graphical method (your calculator or you can use foop?ot.com) to solve the cubic equation (obviously you can't show work for that step, so get the equation, and tell me what method you used to solve for x)

Explanation / Answer

d. NH3 (aq) + H2O (l) ------> NH4+ (aq) + OH- (aq)

And kb = [NH4+ ][OH-]/[NH3]

We need to find concentrations of[NH4+ ][OH-] and [NH3] at equilibrium

Based on the the reaction conditions given would be:

NH3 (aq) + H2O (l) ------> NH4+ (aq) + OH- (aq) we can setup the ICE table as follows:

Initial (I) 0.25 M 0 0

Change -x +x +x

Equi(E) (0.25M -x) x x

kb of NH3 = 1.8 x 10-5

kb = x.x/(0.25-x)

1.8 x 10-5= x2/(0.25-x)

Since NH3 is a weak base, x is so small that it can be ignored in the denominator.

=> 1.8 x 10-5= x2/0.25 => x2 = 0.25 x 1.8 x 10-5 = 0.45 x 10-5 = 4.5 x 10-6

x= sqrt (4.5 x 10-6) = 2.12 x 10-3

the value of x is2.12 x 10-3 which is much smaller than 0.25M, so the assumption made in case of x is correct.

x= [OH-] = [NH4+] = 2.12 x 10-3 M

[NH3] = 0.25 - 2.12 x 10-3 = 0.248 M

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