Question

what is the ph of a 0.20 M solution of NH4Cl? [Kb(NH3) = 1.8 x 10^-5]

Answer 1

"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"

Answer 2

Answer:

• The pH of a 0.2M solution of $NH_4Cl$ = $4.98$

Explanation:

$NH_4CL$ is formed by $HCL + NH_3$

=> $NH_4CL -----> NH_4^+ +CL^-$

=> $NH_4^+ + H_2O -----> NH_3 + H_3O^+$

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

$pH = -log(1.05*10{−5})\\\\pH = 4.98$

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