a. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M NaOH (aq), at what

Question

a. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M NaOH (aq), at what volume of NaOH (aq) should the equivalence point be reached and why? If an additional 3.0 mL of 0.1 M NaOH (aq) is then added, what is the expected pH of the final solution?

b. What is the initial pH expected for a 0.1 M solution of acetic acid? For the titration of 25.0 mL of 0.1 M acetic acid with 0.1 M NaOH (aq), at what volume of NaOH (aq) is the equivalence point reached? Is the pH at the equivalence point greater than, less than or the same as in problem #a above? Explain.

c. What is the initial pH expected for a 0.1 M solution of phosphoric acid (H3PO4)? For 25.0 mL of 0.1M H3PO4 (aq), what volume of 0.1 M NaOH (aq) is required to fully titrate all three protons to their end points?

d. In the titration of a weak acid with a strong base, how is the half equivalence point determined and what is its significance? How are the pKa and Ka of the weak acid determined from the half equivalence point?

Explanation / Answer

a. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M NaOH (aq), at what volume of NaOH (aq) should the equivalence point be reached and why? If an additional 3.0 mL of 0.1 M NaOH (aq) is then added, what is the expected pH of the final solution?

Answer: The equivalence point is reached when the number of moles of HCl will become equal to moles of NaOH

Mole = Molarity X volume

So moles of HCl = Moles of NaOH

(Molarity x volume )HCl = (Molarity X volume) NaOH

0.1 X 25 = 0.1 X Volume

Volume of NaOH = 25mL (point of equivalence)

If we add extra 3mL of NaOH then the pH of solution can be calculated as

pOH = 14-pH

pOH = -log[OH-]

[OH-] = Moles of OH- / Volume of solution

Moles of OH- = Molarity X volume = 3 X 0.1/ 1000 = 3 X 10^-4 moles

(multiplied with 1000 to convert mL to L)

[OH-] = 3 X 10^-4 moles / (volume of acid + base) = 3 X 10^-4 moles X1000 / 53 mL

[OH-] = 0.00567 M

pOH = 2.25

pH = 14 - 2.25 = 11.75

b. What is the initial pH expected for a 0.1 M solution of acetic acid? For the titration of 25.0 mL of 0.1 M acetic acid with 0.1 M NaOH (aq), at what volume of NaOH (aq) is the equivalence point reached? Is the pH at the equivalence point greater than, less than or the same as in problem #a above? Explain.

Answer : The intial pH will depend upon the Ka of acetic acid as it is weak acid and will dissociate very low

CH3COOH --> CH3COO- + H+

Ka = 1.75 x 10^-5 = [H+] [CH3COO-] / [CH3COOH]

[H+] = (Ka X 0.1)^1/2 = 0.243 X 10^-3

pH = 3.61

(ii) The equivalence point is reached when the number of moles of CH3COOH will become equal to moles of NaOH

Mole = Molarity X volume

So moles of CH3COOH = Moles of NaOH

(Molarity x volume )CH3COOH = (Molarity X volume) NaOH

0.1 X 25 = 0.1 X Volume

Volume of NaOH = 25mL (point of equivalence)

(iii) The equivalence point leads to formation of a salt of weak acid and strong base

CH3COONa, which will hydrolyze and will make solution basic, so the pH of solution will be mor than 7.

c. What is the initial pH expected for a 0.1 M solution of phosphoric acid (H3PO4)? For 25.0 mL of 0.1M H3PO4 (aq), what volume of 0.1 M NaOH (aq) is required to fully titrate all three protons to their end points?

Answer: This is triprotic acid, however we wll assume that the major portion of H+ are due to first dissociation and the other dissociation constants are very low

Ka1 = 7.5 x 10^-3

So H+ = (Ka X concentration ) ^1/2 = ( 7.5 x 10^-3 X 0.1)^1/2 = 0.875 X 10^-2

pH = -log[H+] = 2.05

d. In the titration of a weak acid with a strong base, how is the half equivalence point determined and what is its significance? How are the pKa and Ka of the weak acid determined from the half equivalence point?

Answer: The half equivalence point is determined by plotting a graph between pH and volume of NaOH.

The half equivalence point is when the pKa of the weak acid becomes equal to the pH of solution.

Or at half equivalence point the pH noted for solution should be equal to the pKa of weak acid.

From titration curve, the middle of gradual curve gives the half equivalence point or pKa of acid

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