Titration: Strong Acid/Base

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.

Sample Solution