As you know, ammonia acts as a weak base in aqueous solution, so right from the start, you should expect the #"pH"# of the solution to be #> 7#.
The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant, #K_b#.
Now, ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take #x# #"M"# to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain
The solution will also contain
This means that the expression of the base dissociation constant will now take the form
which is equal to
Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation
because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left. You now have
Rearrange and solve for #x# to get
Since #x# #"M"# represents the equilibrium concentration of the hydroxide anions, you can say that
Now, an aqueous solution at #25^@"C"# has
Since
you can say that the #"pH"# of the solution is given by
Plug in your value to find
The answer is rounded to two decimal places because you have two sig figs for the molarity of the solution. doyin o. What is the pH of a 0.100 M solution of NaHCO3 at 25∘C, given that the Ka of H2CO3 is 4.3×10−7? 1 Expert Answer |