Problems by Topic - The Common Ion Effect and Buffers Solve an equilibrium probl

Question

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.

Explanation / Answer

(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

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