Why, when an ammonia solution is titrated with a strong acid, the #pH# remains relatively constant up to the addition of a stoichiometric quantity of the acid?

1 Answer
Mar 26, 2017

Answer:

Because you have described formation of a buffer solution.........

Explanation:

A #"buffer solution"# is a mixture of a weak acid and its conjugate base in APPRECIABLE quantities, and buffers act to resist GROSS changes in #pH#.

This is seen from the well-known buffer equation, the which we may represent as:

#pH=pK_a+log_10{[[A^-]]/[[HA]]}#

Where #[A^-]# is the conjugate base. And #[HA]# is its conjugate acid.

In this ammoniacal scenario, the free base is the uncharged species, the ammonia, and the ammonium acid is charged:

#pH=pK_a+log_10{[[NH_3]]/[[NH_4^+]]}#

Many students have difficulties with the logarithmic function. I do not suggest that you do. But I can encourage you and others not to be intimidated when you see a #log# function. When I write #log_ab=c#, I ask to what power I raise the base #a# to get #c#. Here, #a^c=b#. And thus #log_(10)10=1,# #log_(10)100=2, ##log_(10)10^(-1)=-1 #. And #log_(10)1=0#.

Back in the day, before the advent of electronic calculators, students (and scientists, and engineers, statisticians, and economists) would routinely use log tables to do calculations that a calculator would these days render trivial.