Calculate the change in pH when 9.00 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).

Question

anonymous · Answer

pOH = pKb + log [NH4+]/ [NH3]

pKb of ammonia = 4.74

initial pOH = 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

delta pH = 9.26 - 9.23 =0.03



adding OH- the reaction is
NH4+ + OH- = NH3 + H2O
the moles of NH4+ will decrease and the moles of NH3 will increase
I can not answer your second question because thr molar concentration of NaOH is not given