Respuesta :
Answer: The equilibrium concentration of HCl is [tex]2.26\times 10^{-3}M[/tex]
Explanation:
We are given:
Moles of [tex]NH_4Cl(s)[/tex] = 0.564 moles
Volume of vessel = 1.00 L
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}[/tex]
Molarity of [tex]NH_4Cl=\frac{0.564}{1}=0.564M[/tex]
The given chemical equation follows:
[tex]NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)[/tex]
Initial: 0.564
At eqllm: 0.564-x x x
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=[NH_3][HCl][/tex]
The concentration of pure solid and pure liquid is taken as 1.
We are given:
[tex]K_c=5.10\times 10^{-6}[/tex]
Putting values in above equation, we get:
[tex]5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M[/tex]
Negative sign is neglected because concentration cannot be negative.
So, [tex][HCl]=2.26\times 10^{-3}M[/tex]
Hence, the equilibrium concentration of HCl is [tex]2.26\times 10^{-3}M[/tex]
