Question

1) Calculate the change in pH when 3.00ml of .1M HCl(aq) is added to 100 ml of a buffer solution that is .1 M in NH3(aq) and .1M in NH4Cl(aq)

2)Calculate the change in pH when 3.0ml of .1 M NaOH (aq) is added to the original buffer solution

Answer 1

PKb of ammonia = 4.74

From Henderson Equaton

pOH = pKb + log [NH4+]/ [NH3] =4.74

moles of NH3= 0.1*100/1000= 0.1*0.1= 0.01

HCl supplements H+ and moles of HCl added= 0.1*3/1000=0.0003

The reaction is NH3+H+ ---> NH4+. There will be an increase in [NH4+] and decrease in [NH3+]

therefore [NH4+] =0.01+0.0003 =0.0103 [NH3] = 0.01-0.0003=0.0097

pOH= 4.74+log(0.0103/0.0097)=4.766, pH= 14-4.766=9.234

b) The addition of NaOH suppliments OH- and the reaction is NH3+ ,NH4⁺ +OH^- ---> NH₃+ H2O

This will increase NH3 and decrease NH4+

Moles of OH- supplimented from NaOH= 0.1*3/1000=0.0003

[NH3] =0.01+0.0003 =0.0103 [NH4+]=0.01-0.0003=0.0097

pOH= 4.74+log (0.0097/0.0103)=4.72

pH= 14-4.72=9.28