Question

Calculate the change in ph when 3.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq) and 0.100 m in nh4cl(aq). a list of ionization constants can be found here.

Answer 1

The change in pH is calculated by:

pOH = Protein kinase B + log [NH4+]/ [NH3] 

Protein kinase B of ammonia = 4.74 

initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74 
pH = 14 - 4.74=9.26 

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100 
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300 

NH3 + H+ = NH4+ 
moles NH3 = 0.0100 - 0.000300=0.00970 
moles NH4+ = 0.0100 + 0.000300=0.0103 

pOH = 4.74 + log 0.0103/ 0.00970= 4.77 
oH = 14 - 4.77 = 9.23 

the change is  = 9.26 - 9.23 =0.03