An #0.500*L# volume of ammonia/ammonium buffer has #pH=9.15#. If #[NH_3]=0.10*mol*L^-1#, what MASS of ammonium chloride does this buffer contain?

1 Answer
Mar 25, 2017

Answer:

We use the buffer equation, #pH=pK_a+log_10{[[NH_3]]/[[NH_4^+]]}#

Explanation:

Now for #"ammonium ion"# in water #pK_a=9.24#

And thus, #9.15=9.24+log_10{[[NH_3]]/[[NH_4^+]]}#

Clearly, #log_10{[[NH_3]]/[[NH_4^+]]}=-0.09#

And so, #[[NH_3]]/[[NH_4^+]]=10^(-0.09)=0.813#

And given that #[NH_3]=0.10*mol*L^-1#, #[NH_4^+]=0.123*mol*L^-1#.

Because the volume of the solution was specified to be #0.500*L#, there were #0.500*Lxx0.123*mol*L^-1=0.0615*mol#, i,e, #0.0615*molxx53.49*g*mol^-1=3.29*g#.

For the derivation of the #"buffer equation"#, see here for details.