Chemistry: Acids and Bases
Equilibrium, pH, buffers, titrations, and acid-base theories
Chemistry: Acids and Bases
Equilibrium, pH, buffers, titrations, and acid-base theories
Chemistry - Grade advanced
- 1
A 0.0250 M solution of hydrofluoric acid, HF, has Ka = 6.8 x 10^-4. Calculate the pH of the solution using an equilibrium expression.
Do not assume x is negligible unless you check the percent ionization.
For HF ⇌ H+ + F-, Ka = x^2/(0.0250 - x). Solving gives x = [H+] = about 0.00379 M, so pH = -log(0.00379) = 2.42. The pH of the solution is about 2.42. - 2
A weak base B has Kb = 1.6 x 10^-5. What is the pH of a 0.100 M solution of B?
Find [OH-] first, then convert pOH to pH.
For B + H2O ⇌ BH+ + OH-, Kb = x^2/(0.100 - x). Using the weak base approximation, x = sqrt((1.6 x 10^-5)(0.100)) = 0.00126 M OH-. The pOH is 2.90, so the pH is 14.00 - 2.90 = 11.10. - 3
Calculate the pH of a buffer made by mixing 0.200 mol of acetic acid, HC2H3O2, and 0.300 mol of sodium acetate, NaC2H3O2, in enough water to make 1.00 L of solution. The Ka of acetic acid is 1.8 x 10^-5.
Use pH = pKa + log([base]/[acid]). Because both are in the same final volume, moles may be used in the ratio.
The pKa is -log(1.8 x 10^-5) = 4.74. Using the Henderson-Hasselbalch equation, pH = 4.74 + log(0.300/0.200) = 4.74 + 0.176 = 4.92. The buffer pH is about 4.92. - 4
A buffer contains 0.250 M NH3 and 0.150 M NH4Cl. The Kb of NH3 is 1.8 x 10^-5. Calculate the pH of the buffer.
The pKa of NH4+ is 14.00 - pKb. Since pKb = -log(1.8 x 10^-5) = 4.74, pKa = 9.26. The pH is 9.26 + log(0.250/0.150) = 9.26 + 0.222 = 9.48. The buffer pH is about 9.48. - 5
A 25.00 mL sample of 0.1000 M HCl is titrated with 0.1000 M NaOH. Calculate the pH after 10.00 mL of NaOH has been added.
Before the equivalence point in a strong acid-strong base titration, calculate the excess strong acid.
Initial moles of HCl are 0.02500 L x 0.1000 M = 0.002500 mol. Moles of NaOH added are 0.01000 L x 0.1000 M = 0.001000 mol. Excess H+ is 0.001500 mol in a total volume of 0.03500 L, so [H+] = 0.04286 M. The pH is -log(0.04286) = 1.37. - 6
A 25.00 mL sample of 0.1000 M acetic acid, Ka = 1.8 x 10^-5, is titrated with 0.1000 M NaOH. Calculate the pH at the half-equivalence point.
For a weak acid titrated with strong base, the half-equivalence point has pH equal to pKa.
At the half-equivalence point, half of the weak acid has been converted into its conjugate base, so [A-] = [HA]. The Henderson-Hasselbalch equation becomes pH = pKa. Since pKa = -log(1.8 x 10^-5) = 4.74, the pH is 4.74. - 7
A 40.00 mL sample of 0.200 M HNO3 is mixed with 60.00 mL of 0.150 M KOH. Determine whether the final solution is acidic, basic, or neutral, and calculate the final pH.
Moles of HNO3 are 0.04000 L x 0.200 M = 0.00800 mol. Moles of KOH are 0.06000 L x 0.150 M = 0.00900 mol. KOH is in excess by 0.00100 mol. The total volume is 0.10000 L, so [OH-] = 0.0100 M. The pOH is 2.00 and the pH is 12.00, so the final solution is basic. - 8
Rank the following 0.10 M acid solutions from lowest pH to highest pH: HCl, HF, CH3COOH, and HCN. Use these Ka values: HF = 6.8 x 10^-4, CH3COOH = 1.8 x 10^-5, HCN = 6.2 x 10^-10.
For acids at the same concentration, stronger acid means larger Ka and lower pH.
The order from lowest pH to highest pH is HCl, HF, CH3COOH, HCN. HCl is a strong acid and dissociates completely, giving the greatest [H+]. Among the weak acids, the larger Ka produces more H+ and therefore a lower pH. - 9
Identify the acid, base, conjugate acid, and conjugate base in this reaction: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq).
NH3 is the base because it accepts a proton. H2O is the acid because it donates a proton. NH4+ is the conjugate acid of NH3, and OH- is the conjugate base of H2O. - 10
In the reaction BF3 + NH3 → F3B-NH3, identify the Lewis acid and the Lewis base. Explain your choice.
A Lewis acid accepts an electron pair, and a Lewis base donates an electron pair.
BF3 is the Lewis acid because boron can accept an electron pair. NH3 is the Lewis base because nitrogen donates a lone pair to form the coordinate covalent bond in F3B-NH3. - 11
A salt solution contains 0.100 M NaF. The Ka of HF is 6.8 x 10^-4. Calculate the pH of the solution, assuming ideal behavior.
The fluoride ion is the conjugate base of a weak acid, so it reacts with water to produce OH-.
F- is a weak base with Kb = Kw/Ka = (1.0 x 10^-14)/(6.8 x 10^-4) = 1.47 x 10^-11. For F- + H2O ⇌ HF + OH-, [OH-] = sqrt((1.47 x 10^-11)(0.100)) = 1.21 x 10^-6 M. The pOH is 5.92, so the pH is 8.08. The solution is slightly basic. - 12
Phosphoric acid, H3PO4, is triprotic with Ka1 = 7.1 x 10^-3, Ka2 = 6.3 x 10^-8, and Ka3 = 4.2 x 10^-13. Explain why the first dissociation dominates the pH of a typical H3PO4 solution.
The first dissociation dominates because Ka1 is many orders of magnitude larger than Ka2 and Ka3. This means H3PO4 loses its first proton much more readily than H2PO4- loses the second proton or HPO4^2- loses the third proton. As a result, most of the hydrogen ions come from the first dissociation. - 13
A titration curve begins at pH 2.9, has a buffer region, reaches equivalence at pH 8.7, and levels off above pH 12. Identify the most likely type of titration and justify your answer.
The pH at the equivalence point is the key feature.
This is most likely a weak acid titrated with a strong base. The initial pH is acidic but not extremely low, the buffer region indicates a weak acid-conjugate base mixture, and the equivalence point is above 7 because the conjugate base of the weak acid makes the solution basic. - 14
A solution has [H+] = 3.2 x 10^-9 M at 25°C. Calculate [OH-], pH, and pOH.
At 25°C, Kw = [H+][OH-] = 1.0 x 10^-14. Therefore [OH-] = (1.0 x 10^-14)/(3.2 x 10^-9) = 3.1 x 10^-6 M. The pH is -log(3.2 x 10^-9) = 8.49, and the pOH is 14.00 - 8.49 = 5.51. - 15
A student adds a small amount of strong acid to a buffer containing HA and A-. Explain, using Le Châtelier's principle, how the buffer resists a large change in pH.
Focus on the reaction H+ + A- → HA.
When strong acid is added, it increases the concentration of H+. The conjugate base A- reacts with the added H+ to form HA, which removes much of the added H+ from solution. By shifting toward HA, the buffer reduces the change in [H+] and resists a large decrease in pH.