What is the pH of a 0.85 M NH_4Cl solution?

1 Answer
May 26, 2016

pH = 4.66

Explanation:

Ammonium chloride is a soluble salt that ionizes completely when dissolved in water.

" "NH_4Cl -> NH_4^++ Cl^-

Cl^(-)is a spectator ion that does not affect the pH of the medium.

On the other hand, NH_4^+is the conjugate species of the weak base\ NH_3\ therefore, it is a weak acid stronger than water. It dissociates in water according to the following equation:

" "NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+

(NH_4^+//NH_3) are conjugate acid base pairs, and for conjugate acid base pairs:

" "K_axxK_b=K_w

Knowing the\ K_b\ value for \ NH_3 (1.8xx10^-5), the \ K_a\ value for NH_4^+ is determined.

K_a =K_w/K_b

K_a =(1.00xx10^-14)/( 1.80xx10^-5)

K_a =5.56xx10^-10

Use the ICE table to figure out the concentration of each species present at equilibrium.

" "ul(NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+)
\ "I "0.85 M" " - " " -
"C " -x" " +x" "+x
"E " ul( 0.85-x" "x" " x" "

K_a= ([NH_3]xx[ H_3O^+] )/([NH_4^+])

K_a= (x.x)/((0.85-x))=5.56xx10^-10

(0.85-x)~= 0.85 since x is too small compared to 0.85

(x^2)/(0.85)=5.56xx10^-10

x= sqrt(0.85xx5.56xx10^-10)

x~=2.2xx10^-5\ M

x=[H_3O^+]

pH=-log\ (2.2xx10^-5)

pH = 4.66