Part A : A 14.0 mL sample of vinegar, which is an aqueous solution of acetic aci

Question

Part A : A 14.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 24.5 mL of a 0.650 M solution of NaOH to reach the endpoint in a titration:

CH3COOH(aq)+NaOH(aq)?NaCH3COO(aq)+H2O(l)

What is the molarity of the acetic acid solution? (Express the concentration to three significant figures and include the appropriate units.)

Part B: Calculate the pH of each solution given the following [H3O+] or [OH?] values.

[H3O+] = 3.0×10?4 M

[H3O+] = 1.0×10?8 M

[OH?] = 5.0×10?5 M

[OH?] = 2.0×10?11 M

[H3O+] = 7.0×10?8 M

[OH?] = 7.5×10?4 M

Express your answer using two decimal places.

Part C: Solution X has a pH of 10.1, and solution Y has a pH of 7.6.

1. Which solution is more acidic?  

A.solution Y OR B. solution X

2. What is the [H3O+] in solution X?

Express your answer to one significant figure and include the appropriate units.

3. What is the [H3O+] in solution Y?

Express your answer to one significant figure and include the appropriate units.

4. What is the [OH?] in solution X?

Express your answer to tone significant figure and include the appropriate units.

5. What is the [OH?] in solution Y?

Express your answer to one significant figure and include the appropriate units.

Explanation / Answer

art A : A 14.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 24.5 mL of a 0.650 M solution of NaOH to reach the endpoint in a titration:

moles = molarity* volume in liters, 1000ml=1 L, moles of NaOH used= 0.650*24.5/1000 =0.016 moles

the reaction that takes place is

CH3COOH(aq)+NaOH(aq)?NaCH3COO(aq)+H2O(l)

1 mole of NaOH requires 1 mole of acetic acid for complete neutralization

0.016 moles of NaOH requires 0.016 moles of acetic acid.

volume of vinegar= 14 ml= 14/1000 L, concentration of acetic acid in vinegr= 0.016/(14/1000)= 1.14M

2.

pH= -log [H3O+] pH= 14-pOH, pOH= -log [OH-]

[H3O+] = 3.0×10?4 M, pH= -log (3*10-4)= 3,5

[H3O+] = 1.0×10?8 M, pH= -log (1*10-8)= 8

[OH?] = 5.0×10?5 M, pOH= -log (5*10-5)= 4.3, pH= 14-4.3= 9.7

[OH?] = 2.0×10?11 M, pOH= -log (2*10-11)= 10.7, pH= 14-10.7= 3.3

[H3O+] = 7.0×10?8 M, pH= -log (7*10-8)= 7.15

[OH?] = 7.5×10?4 M, pOH= -log (7.5*10-4)= 3.12, pH= 14-3.12= 10.88

3. if PH lies between 0 to 7, the solution is acidic, the acidity decreases as pH increases since pH= -log [H3O+], lower the pH, higher the acidity and higher the pH lower the acidity. if PH>7, the solution is basic and higher the pH higher the basicity.

Part C: Solution X has a pH of 10.1, and solution Y has a pH of 7.6.

since solution Y has lower pH, it is acidic,

2. What is the [H3O+] in solution X?

pH= -log [H3O+], [H3O+]= 10(-pH)= 10(-10.1)= 7.9*10-11

3. What is the [H3O+] in solution Y?

pH= 7.6, [H3O+]= 10(-7.6)= 2.5*10-8

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