11.49

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604744616
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Joined: Sun Jun 25, 2017 3:00 am

11.49

Postby 604744616 » Tue Aug 01, 2017 2:44 pm

With the last part of this homework problem can someone please explain to me why and how they found those values for [NH^3] and [H^2S]?

Stephanie H
Posts: 20
Joined: Fri Jun 23, 2017 11:39 am

Re: 11.49

Postby Stephanie H » Tue Aug 01, 2017 5:18 pm

To find the equilibrium concentrations of NH3 and H2S,

first write out the equilibrium constant expression: Kc= [NH3]*[H2S] / 1
-I placed a one in the denominator because the concentration of a solid, which in this case is NH4HS, is 1

then make a table to find the missing values:

initial concentration (0.2 mol/2L) (NH3) and 0 (H2S)
change in concentration +x (NH3) and +x (H2S)
equilibrium concentration (0.2+x) (NH3) and (x) (H2S)

now plug in the equilibrium concentrations into Kc to solve for x:

(1.6* 10^-4) = (0.2+x)*(x)

x^2+ 0.2x-(1.6*10^-4)=0

x= 7.98*10^-4 or -0.2..

disregard the negative value because concentrations are always positive

x= 7.98*10^-4

since we know the value of x we can plug it into the equilibrium concentration expressions we found and calculate the final concentrations

NH3: 0.2+ (7.98* 10^-4)= 0.2 mol/L

H2S: x= 8 * 10^-4 mol/L


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