What is the pH of a 0.85 M #NH_4Cl# solution?

1 Answer
May 26, 2016

Answer:

#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#