Question

Use the Henderson–Hasselbalch equation to calculate the pH of each solution:

Part A a solution that is 0.135 M in HClO and 0.165 M in KClO Express your answer using two decimal places. Part B a solution that contains 1.23% C2H5NH2 by mass and 1.30% C2H5NH3Br by mass Part C a solution that is 15.0 g of HC2H3O2 and 15.0 g of NaC2H3O2 in 150.0 mL of solution

Answer 1

A. Henderson–Hasselbalch equation for acidic buffer solution:

pH = pK_a + log[salt]/[acid]

= 7.54 + log 0.165/0.135

= 7.63

B. Molar mass of C₂H₅NH₂ = 45 g/mol

Let 1.23 g C₂H₅NH₂ is present in 1000 mL solution.

Concentration of C₂H₅NH₂ = 1.23 g/(45 g/mol)

= 0.027 M

Molar mass of C₂H₅NH_­Br = 124 g/mol

Let 1.30 g C₂H₅NHBr is present in 1000 mL solution.

Concentration of C₂H₅NHBr = 1.30 g/(124 g/mol)

= 0.010 M

Henderson–Hasselbalch equation for basic buffer solution:

pOH = pK_b + log[salt]/[base]

= 3.37 + log 0.01/0.027

= 2.96

pH = 14 - 2.96

= 11.04

C. First we have to determine the concentration of the components

Number of moles of HC2H3O2 present in 150 mL solution

= 15 g/(60 g/mol)

= 0.25 moles

Concentration of HC2H3O2 = 0.25 moles * 1000/150 mL

= 1.67 M

Number of moles of NaC2H3O2 present in 150 mL solution

= 15 g/(82 g/mol)

= 0.18 moles

Concentration of NaC2H3O2 = 0.18 moles * 1000/150 mL

= 1.22 M

Henderson–Hasselbalch equation for acidic buffer solution:

pH = pK_a + log[salt]/[acid]

= 4.74 + log 1.22/1.67

= 4.60