In: Chemistry
Problems by Topic - The Common Ion Effect and Buffers
Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following.
(1) a solution that is 0.17 M in HCHO2 and 0.11 M in NaCHO2
(2) a solution that is 0.12 M in NH3 and 0.16 M in NH4Cl
2nd set of problems
Part A
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.110 M in HClO and 0.155 M in KClO.
Part B
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.180 M in C2H5NH2 and 0.155 M in C2H5NH3Br.
Part C
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 10.5 g of HC2H3O2 and 13.5 g of NaC2H3O2 in 150.0 mL of solution.
(1) a solution that is 0.17 M in HCHO2 and 0.11 M in NaCHO2
formic acid pKa = 3.75
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
pH = 3.75 + log[0.11/0.17]
pH = 3.56
(2) a solution that is 0.12 M in NH3 and 0.16 M in NH4Cl
pKb of NH3 = 4.74
For Basic buffer
Henderson-Hasselbalch equation
pH = 14 –{ pKb + log[salt/ base]}
pH = 14 –{ 4.74 + log[0.16/ 0.12]}
pH = 9.14
Part A
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.110 M in HClO and 0.155 M in KClO.
HClO pKa = 7.52
pH = pKa + log[salt/acid]
pH = 7.52 + log[0.155/0.110]
pH = 7.67
Part B
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.180 M in C2H5NH2 and 0.155 M in C2H5NH3Br.
pKb of C2H5NH2 =3.37
Henderson-Hasselbalch equation
pH = 14 –{ pKb + log[salt/ base]}
pH = 14 –{ 3.37 + log[0.155/ 0.180]}
pH = 10.69
Part C
Use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 10.5 g of HC2H3O2 and 13.5 g of NaC2H3O2 in 150.0 mL of solution
HC2H3O2 molarity = (10.5 / 60 ) x 1000/150
= 1.17M
NaC2H3O2 molarity = (13.5 / 82) x 1000/ 150
= 1.097 M
acetic acid pKa = 4.74
pH = pKa + log[salt/acid]
pH = 4.74 + log[1.097 / 1.17]
pH = 4.71