Question

What is the pH of a buffer made by mixing 65.0 mL of 0.50 M aqueous ammonia with 50.0 mL 0.50 M ammonium nitrate?

Answer 1

Concentration after mixing = mol of component / (total volume)

M(NH4+) after mixing = M(NH4+)*V(NH4+)/(total volume)

M(NH4+) after mixing = 0.5 M*65.0 mL/(65.0+50.0)mL

M(NH4+) after mixing = 0.2826 M

Concentration after mixing = mol of component / (total volume)

M(NH3) after mixing = M(NH3)*V(NH3)/(total volume)

M(NH3) after mixing = 0.5 M*50.0 mL/(50.0+65.0)mL

M(NH3) after mixing = 0.2174 M

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.2174/0.2826}

= 4.63

use:

PH = 14 - pOH

= 14 - 4.63

= 9.37

Answer: 9.37