Respuesta :

pultimax

"NH4+ <----> NH3 + H+  

The constant of this equilibrium is: K = Kw / Kb = 1 x 10^-14 / 1.8 x 10^-5 =5.56 x 10^-10  

5.56 x 10^-10 = x^2 / 0.20-x  

x = [H+] =1.1 x 10^-5 M  

pH = 5.0"

okpalawalter8

Answer:

  • The pH of a 0.2M solution of [tex]NH_4Cl[/tex] = [tex]4.98[/tex]

Explanation:

[tex]NH_4CL[/tex] is formed by [tex]HCL + NH_3[/tex]

=> [tex]NH_4CL -----> NH_4^+ +CL^-[/tex]

=> [tex]NH_4^+ + H_2O -----> NH_3 + H_3O^+[/tex]

[tex]H^+ = \sqrt{\frac{kw*C}{k_b}} \\\\= \sqrt{\frac{10^{-14}*0.2}{1.8*10^{-5}}}\\\\= 1.05*10{−5}[/tex]

[tex]pH = -log(1.05*10{−5})\\\\pH = 4.98[/tex]

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