Use the Henderson–Hasselbalch equation to calculate the pH of each solution: Par
ID: 1036950 • Letter: U
Question
Use the Henderson–Hasselbalch equation to calculate the pH of each solution:
Part A
Part complete
a solution that is 0.115 M in HClO and 0.175 M in KClO
Express your answer using two decimal places.
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Part B
Part complete
a solution that contains 1.46% C2H5NH2 by mass and 1.00% C2H5NH3Br by mass
Express your answer using two decimal places.
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Part C
Part complete
a solution that is 10.0 g of HC2H3O2 and 13.5 g of NaC2H3O2 in 150.0 mL of solution
Use the Henderson–Hasselbalch equation to calculate the pH of each solution:
Part A
Part complete
a solution that is 0.115 M in HClO and 0.175 M in KClO
Express your answer using two decimal places.
SubmitPrevious AnswersRequest Answer
Part B
Part complete
a solution that contains 1.46% C2H5NH2 by mass and 1.00% C2H5NH3Br by mass
Express your answer using two decimal places.
pH =SubmitPrevious AnswersRequest Answer
Part C
Part complete
a solution that is 10.0 g of HC2H3O2 and 13.5 g of NaC2H3O2 in 150.0 mL of solution
pH =Explanation / Answer
Part-A
[HClO] = 0.115M
[KClO] =0.175M
Pka of HClO= 7.54
PH= PKa + log[salt]/[acid]
PH= 7.54 + log(0.175/0.115)
PH= 7.72
B)
C2H5NH2= 1.46% by mass
molar massof C2H5NH2 =45.08 gram/mole
number of molesof C2H5NH2 = 1.46/45.08 = 0.0324 moles
C2H5NH3Br= 1.00%
molar mass of C2H5NH3Br = 125.99 gram/mole]
number of moles of C2H5NH3Br= 1.00/125.99 = 0.00794 moles
PKb of CH3CH2NH2 = 3.25
POH = Pkb + log[salt[/[acid]
POH= 3.25 + log(0.00794/0.0324)
POH= 2.64
PH+POH= 14
PH+ 14-POH
PH= 14-2.64
PH= 11.36
C)mass of acetic acid = 10.0grams
molar mass of cetic acid = 60.0gram/mole
number of moles of acetic acid= 10.0/60.0= 0.167 moles
mass of sodium acetate = 13.5 gramns
molar mass of Sodium acetate= 82.0gram/mole
nubmer of moles of Sodium acetate = 13.5/82.0 = 0.165 moes
Pka of acetic acid = 4.75
PH= Pka + log[salt]/[acid]
PH= 4.75 + log(0.165/0.167)
PH= 4.74
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