What is the equilibrium concentration of #"HCl"#?

The equilibrium constant, #K_c#, for the following reaction is #5.10xx10^(-6)# at #"548 K"#.

#"NH"_4"Cl"(s) rightleftharpoons "NH"_3(g) + "HCl"(g)#

Calculate the equilibrium concentration of #"HCl"# when #"0.471 moles"# of #"NH"_4"Cl"(s)# are introduced into a #"1.00 L"# vessel at #"548 K"#.

1 Answer
Feb 19, 2018

#[HCl(g)]=[NH_3(g)]=2.26xx10^-3*mol*L^-1#

Explanation:

We gots...

#NH_4Cl(s)rightleftharpoonsNH_3(g) + HCl(g)#...

...for which #K_(eq)=[NH_3(g)][HCl(g)]=5.10xx10^-6# at #548*K#.

And the reactant ammonium chloride does not appear in this equation, given that a SOLID cannot express a concentration....that they give a mass here is in order to distract you...

Now if we set #x=[NH_3(g)]=[HCl(g)]#...

...then #x^2=5.10xx10^-6#, and #x=sqrt(5.10xx10^-6)=2.26xx10^-3*mol*L^-1#..