An ammonium chloride/ammonia buffer solution is to be prepared with a pH of 10.25. Calculate the [NH3(aq)] / [NH4+(aq)] ratio necessary for this buffer solution. The Ka for ammonia = 1.8x10^-5.
Kb of NH3 = 1.8*10^-5
1st find Ka of NH4+
use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/1.8*10^-5
Ka = 5.556*10^-10
pKa = - log (Ka)
= - log(5.556*10^-10)
= 9.255
use:
pH = pKa + log {[conjugate base]/[acid]}
10.25 = 9.255+log {[NH3]/[NH4+]}
log {[NH3]/[NH4+]} = 0.9947
[NH3]/[NH4+] = 9.879
Answer: 9.88
An ammonium chloride/ammonia buffer solution is to be prepared with a pH of 10.25. Calculate the...