Use the Henderson–Hasselbalch equation to calculate the pH of each solution: Par
ID: 1051024 • Letter: U
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
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
Explanation / Answer
A. Henderson–Hasselbalch equation for acidic buffer solution:
pH = pKa + 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 C2H5NHBr = 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 = pKb + 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 = pKa + log[salt]/[acid]
= 4.74 + log 1.22/1.67
= 4.60
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.