Titration: Strong Acid/Base

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1. 10 mL of M HCl solution is mixed with 150 mL of water to make a new HCl solution. What is the pH of the resultant solution? Justify your answer.

2. Let’s prepare an H2SO4 solution by dissolving 4.9 grams of H2SO4 in 100 mL of water.
a. How many H2SO4 molecules are in this solution?
b. What is the molarity of the solution?
c. What is the pH of the solution?
d. What is the pH of the solution if 3.0 grams of NaOH are added to the solution?
Note: Sulfuric acid (H2SO4) is a diprotic acid. The dissociation constant of the first proton (H+) is significantly larger than the second H+ dissociation. So, the pH of such a solution is mainly contributed by the first proton dissociation, while the second proton dissociation will provide some modification to its pH value. If you haven’t learned polyprotic acids yet, consider one H+ dissociation is enough to be awarded full credit.

Sample Solution

Problem 1:

Given:

  • 10 mL of 1 M HCl solution
  • 150 mL of water

Find:

  • pH of the resultant solution

Solution:

  1. Calculate the total volume of the final solution: Final volume = 10 mL HCl + 150 mL water = 160 mL

  2. Calculate the total moles of H+ ions from the HCl: Moles of H+ = Volume of HCl * Molarity of HCl = 0.010 L * 1 M = 0.010 mol

  3. Calculate the final concentration of H+ ions: Concentration of H+ = Moles of H+ / Final volume = 0.010 mol / 0.160 L = 0.0625 M

  4. Calculate the pH: pH = -log[H+] = -log(0.0625) ≈ 1.20 (rounded to two significant figures)

Justification:

  • We assumed complete dissociation of HCl in water, which is a valid assumption for a strong acid like HCl.
  • We used the definition of pH and the calculated concentration of H+ ions to determine the pH.

Therefore, the pH of the resultant solution is approximately 1.20.

Problem 2:

Given:

  • 4.9 grams of H2SO4
  • 100 mL of water (0.1 L)

Find:

a. Number of H2SO4 molecules b. Molarity of H2SO4 solution c. pH of H2SO4 solution d. pH of solution after adding 3.0 grams of NaOH

Solution:

a. Number of H2SO4 molecules:

  1. Calculate moles of H2SO4: Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4 = 4.9 g / 98.09 g/mol ≈ 0.050 mol
  2. Calculate number of molecules: Number of molecules = Moles of H2SO4 * Avogadro’s number ≈ 0.050 mol * 6.022 x 10^23 molecules/mol ≈ 3.01 x 10^22 molecules

b. Molarity of H2SO4 solution:

  • Molarity already calculated in step 1a: 0.050 M

c. pH of H2SO4 solution (considering only first dissociation):

  1. H2SO4 is a diprotic acid, but we’ll consider only the first dissociation for now.
  2. Assume complete dissociation of H2SO4 to H+ and HSO4^-: H2SO4 -> H+ + HSO4^-
  3. Therefore, [H+] = [HSO4^-] = 0.050 M
  4. Calculate pH assuming 1:1 ionization (which underestimates the actual acidity): pH = -log[H+] = -log(0.050) ≈ 1.30

d. pH of solution after adding 3.0 grams of NaOH:

  1. Calculate moles of NaOH: Moles of NaOH = Mass of NaOH / Molar mass of NaOH = 3.0 g / 40.00 g/mol ≈ 0.075 mol
  2. NaOH reacts with H+ from H2SO4: NaOH + H+ -> H2O + Na+
  3. H+ is the limiting reagent since we have less of it (0.050 mol) than NaOH (0.075 mol).
  4. Therefore, all H+ from H2SO4 reacts, leaving 0.050 mol HSO4^- and 0.025 mol unreacted NaOH (excess base).
  5. HSO4^- partially dissociates, but its Ka2 is much smaller than Ka1 (Ka1 << Ka2), so we can approximate using Ka1 only: HSO4^- -> H+ + SO4^2- Ka1 = [H+][SO4^2-] / [HSO4^-] = 1.2 x 10^-2
  6. Since [HSO4^-] = 0.050 mol and [SO4^2-] is initially 0, we can use the approximation H+: x * x / (0.

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