Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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.

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.

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

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