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.

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