Question

Calculate the change in pH when 7.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH, (aq) and 0.100 M in NH, Cl(aq). Consult the table of ionization constants as needed. ApH = Calculate the change in pH when 7.00 mL of 0.100 M NaOH is added to the original buffer solution. ApH =

Answer 1

The number of millimoles of NH, (aq) is calculated as shown below: Number of millimoles = Molarity x Volume( in mL) = 0.100 Mx100mL =10 mmoles The number of millimoles of NH,Cl(aqis calculated as shown below: Number of millimoles = Molarity x Volume (in mL) = 0.100 Mx100 mL =10 mmoles The number of millimoles of HCl(aq) is calculated as shown below: Number of millimoles = Molarity x Volume( in mL) = 0.100 Mx 7 ml = 0.7 mmoles When HCl is added to the buffer solution then the change in buffer is:

When HCl is added to the buffer solution then the change in buffer is: = H2O NH, (aq) Intitial 10 mmoles Change – 0.7 mmoles Final 9.3 mmoles + HCl(aq) 0.7 mmoles -0.7 mmoles NH,Cl(aq) + 10 mmoles +0.7 mmoles 10.7mmoles The pH before addition is calculated as shown below: [salt] | 14-pH =pk +lo (base) 10mmoles =pKo + log S 10 mmoles 14-pH, = PK, ...... (1) The pH after addition of HCl is calculated as shown below: 14-pH, = pK, +log [base] 10.7 mmoles =pKo + log 9.3 mmoles 14-pH, = pK+0.061 Substitute the value of pK, as 14- pH, in the above equation as shown below:

14-pH =14-pH +0.061 ApH = -0.061 The final answer is ApH =-0.061. Similarly, the change in pH with the addition of NaOH is: NH,Cl(aq) Intitial 10 mmoles Change -0.7 mmoles Final 9.3 mmoles + NaOH(aq) 0.7 mmoles -0.7 mmoles = NH3(aq) + H2O 10 mmoles +0.7 mmoles 10.7mmoles The pH after addition of HCl is calculated as shown below: who [salt] 14-pH, = PK, +log [base] 9.3mmoles =pK+log 10.7 mmoles 14-PH3 =pK-0.061 Substitute the value of pK, as 14 - pH, in the above equation as shown below:

| 14-pH =14-pH -0.061 ApH = +0.061 The final answer is ApH = +0.061.