Question

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?

Answer 1

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.](https://socratic.org/questions/how-do-buffers-maintain-ph)