Question

Q) Say you're doing an experiment and it requires a buffer solution with pH 9.15. Calculate how much NH4Cl must be added to 500mL of 0.1M aqueous ammonia to achieve this. (Kb NH3 = 1.8 x 10^-5)?

Answer 1

The buffer equation holds that...

`pH=pK_a+log_10{[[NH_4^+]]/[[NH_3(aq)]]}`

Explanation:

And since, in WATER,,,`K_axxK_b=10^-14`...then `pK_a+pK_b=14`...and if `K_b=1.8xx10^-5`...

`pK_a("ammonium ion")=14-4.74=9.26`

And so we fill in the blanks...

`9.15=9.26+log_10{[[NH_4^+]]/[[NH_3(aq)]]}`

And...

`log_10{[[NH_4^+]]/[[NH_3(aq)]]}=-0.11`

`[[NH_4^+]]/[[NH_3(aq)]]=10^(-0.11)-=0.776`

And we specify that `[NH_3]=0.10*mol*L^-1`...therefore `[NH_4^+]=(0.10xx0.776)*mol*L^-1=0.0776*mol*L^-1`.

But a `0.500*L` volume of buffer was specified....

And therefore `(n_"ammonium chloride")/(0.500*L)=0.0776*mol*L^-1`...

`n_"ammonium chloride"=0.500*Lxx0.0776*mol*L^-1=0.0388*mol`.

`"mass"_"ammonium chloride"=0.0388*molxx53.49*g*mol^-1=2.08*g`