Applications of Aqueous Equilibria

Applications of Aqueous Equilibria • Acid-Base Equilibria • More Acid-Base Equilibria • Solubility of Salts • Formation of Complex Ions

Applications of Aqueous Equilibria • Natural processes that involve these equilibria • weathering of minerals • uptake of nutrients by plants • tooth decay • formation of limestone caverns, stalactites, and stalagmites

Applications of Aqueous Equilibria • Common Ion Problems • Involve solutions that contain a weak acid, HA, and its salt, NaA • Ex: Given a solution containing HF (Ka = 7.2 x 10-4) and its salt, NaF. What happens to the dissociation of HF? • NaF is a strong electrolyte, i.e. it ionizes 100% • NaF ---> Na+ + F-

Applications of Aqueous Equilibria • HF is a weak acid • HF <====> H+ + F- • Therefore, the major species in the solution is Na+, F-, HF, and H2O • F- is a common ion, since it is produced by both HF and NaF.

Applications of Aqueous Equilibria • Apply Le Chatelier’s Principle. The F- present from the ionization of NaF forces the HF to ionize less as the reverse reaction is favored. • The acidity of a solution with a common ion present is less than a solution of HF alone, i.e. the [H+] is less, or the pH is greater.

Applications of Aqueous Equilibria • Buffered Solutions • consists of a solution that contains a weak acid and its salt or a weak base and its salt. • resists changes in pH upon the addition of an acid or base • Ex: Our blood is a buffered solution which can absorb acids or bases without changing its pH, which is important because our cells can only survive in a narrow pH range.

Applications of Aqueous Equilibria • A solution can be buffered to any pH by choosing the appropriate components and concentrations. • Buffered solutions are still only solutions of weak acids or weak bases containing a common ion. The pH calculations are the same as presented before.

Applications of Aqueous Equilibria • How does buffering work? • Buffered solution contains HA and A-. • Add OH- • OH- reacts with the HA: • OH- + HA --> H2O + A- • The OH- gets “eaten up” as it forms water, so with no extra OH- in the system, the pH remains the same.

Applications of Aqueous Equilibria • Add H+ • The H+ reacts with the A- to reform HA: • H+ + A- --> HA • The added H+ gets “eaten up” as it reforms HA, with no extra H+ in the system, the pH remains the same.

Applications of Aqueous Equilibria • Buffered solutions can be formed with a weak base, B, and its salt, BH+. • Add OH- • the OH- will react with the BH+ • OH- + BH+ --> B + H2O • the extra OH- gets “eaten up” by the H+ to form water. • With no extra OH- in the system, the pH stays the same.

Applications of Aqueous Equilibria • Add H+ • The H+ will react with the B to form the salt: • H+ + B --> BH+ • The extra H+ is “eaten up” by the B, with no extra H+ in the system, the pH remains the same.

Applications of Aqueous Equilibria • Henderson-Hasselbach equation • Start with Ka expression: • Ka = [H+][A-]/[HA] • Rearrange to get: [H+] = Ka[HA]/[A-] • Take the negative log of both sides: • pH = pKa - log([HA]/[A-]) or • pH = pKa + log ([A-]/[HA]) or • pH = pKa + log ([base]/[acid]) • Useful for calculating the pH when the [HA]/[A-] ratio is known.

Applications of Aqueous Equilibria • Important characteristics of buffered solutions • they contain large concentrations of a weak acid and the corresponding weak base • HA and A- or • B and BH+

Applications of Aqueous Equilibria • Important characteristics of buffered solutions • Add H+, it will react with the weak base to form a weak acid: • H+ + A- --> HA or • H+ + B --> BH+ • Add OH-, it will react with the weak acid to form a weak base and water: • OH- + HA --> H2O + A- or • OH- + BH+ --> H2O + B

Applications of Aqueous Equilibria • Important characteristics of buffered solutions • the pH of a buffered solution depends on the ratio of the concentrations of the weak acid and the weak base. • While this ratio remains constant, the pH will remain constant.

Applications of Aqueous Equilibria • This ratio will remain constant as long as the concentrations of the buffering materials are large compared to the amounts of H+ or OH- added

Applications of Aqueous Equilibria • Buffer Capacity • the amount of H+ or OH- a buffer can absorb without a significant change in the pH • a buffer with a large capacity can absorb a large amount of H+ or OH- with only a little change in the pH • a buffer with a large capacity contains a large concentration of the weak acid and its salt

Applications of Aqueous Equilibria • pH of a buffered solution • determined by the ratio of [A-]/[HA] • optimal buffering (least change in pH) will occur when [A-] = [HA] • so pH = pKa + log (1) • so pH = pKa + 0 • so pH = pKa • Pick an acid with a Ka closest to the desired pH of the buffered solution.

Applications of Aqueous Equilibria • Titration and pH curves • Titration • used to determine the concentration or amount of an unknown acid or base • uses a solution of known concentration (the titrant) • uses a buret (for precision) • uses an indicator to show the endpoint • can be monitored by plotting pH vs. amount of titrant added

Applications of Aqueous Equilibria • Strong Acid-Strong Base Titrations • millimoles may be used because of the small quantities used in a titration • 1000 millimoles = 1000mmoles = 1 mole • Molarity = moles/liter = mmoles/mL • mmoles = Volume (in mL) x Molarity

Applications of Aqueous Equilibria • Strong Acid-Strong Base Titration • Ex: Titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH • What is the pH at various stages of the titration? • Initially...No NaOH has been added • pH is determined by H+from the 0.200 M HNO3 • pH = - log (0.200) = 0.699 • 50.0 mL x 0.200 M H+ = 10.0 mmol H+ present

Applications of Aqueous Equilibria • Still the beginning of the reaction…10.0 mL of 0.100 M NaOH has been added • The added OH- will neutralize an equivalent amount of H+, I.e., 10.0 ml x 0.100 M or 1.00 mmole. • How much H+ is left? 10.0 mmole - 1.00 mmole = 9.0 mmole H+ • What is the concentration of H+ now? • 9.0 mmole/ (50.0 mL + 10.0 mL) = 0.15 M • pH = -log (0.15) = 0.82…pH is increasing

Applications of Aqueous Equilibria • 20.0 mL of NaOH has been added… • pH = 0.942 • 50.0 mL of NaOH as been added… • pH = 1.301

Applications of Aqueous Equilibria • Strong Acid-Strong Base Titration • At the equivalence point…100.0 mL of NaOH has been added • 100.0 mL x 0.100 M = 10.0 mmole OH- • This is enough OH- to completely react with the H+ in solution. • At the equivalence point, the pH is 7, the solution is neutral

Applications of Aqueous Equilibria • 150.0 mL of NaOH has been added • now OH- is in excess, the pH is determined by the excess OH- • 150.0 mL x 0.100 M = 15.00 mmoles OH- • 15.00 mmoles OH- - 10.0 mmoles H+ = 5.0 mmoles OH- in solution • 5.0 mmoles/ (50.0 + 150.0 mL) = 0.025 M OH-….so [H+] = 4.0 x 10-13…pH = 12.40

Applications of Aqueous Equilibria • 200.0 mL of NaOH has been added • still excess OH- • pH = 12.60

Applications of Aqueous Equilibria • Strong Acid-Strong Base Titration • pH changes very little initially until close to the equivalent point (lots of H+, added OH- doesn’t change the pH much) • Near the equivalence point, there is less H+, so added OH- changes the pH a lot • At the equivalence point, the pH = 7.00 • Just after the equivalence point, added OH- also changes the pH a lot

Applications of Aqueous Equilibria • Weak Acid-Strong Base Titration • Calculating the titration curve is like solving a series of buffer problems. • Involves a stoichiometry problem where the reaction goes to completion and the concentrations of the weak acid and the conjugate base are calculated • Involves an equilibrium problem, calculate pH from this

Applications of Aqueous Equilibria • pH curve for this titration is different before the equivalence point from the strong acid-strong base titration • after the equivalence point, the titration curves are the same

Applications of Aqueous Equilibria • For a weak acid-strong base titration, the pH rises more rapidly in the beginning of the titration, then levels off at the halfway point due to buffering effects. • pH at the equivalence point is higher than for a strong acid-strong base titration

Applications of Aqueous Equilibria • Weak acid-Strong Base titration • The amount of acid determines the equivalence point • The pH value at the equivalence point depends on the acid strength • the weaker the acid, the higher the pH at the equivalence point

Applications of Aqueous Equilibria • Acid-Base Indicators • Two common methods for determining the equivalence point of a titration • Use a pH meter and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point. • Use an acid-base indicator to mark the end point with a change in color.

Applications of Aqueous Equilibria • Acid-Base Indicators • The endpoint (when the indicator changes color) may not be the same as the equivalence point. • An indicator must be chosen based on the acid and base used in the titration so that the endpoint is as close to the equivalence point as possible.

Applications of Aqueous Equilibria • Acid-Base Indicators • Indicators are usually complex molecules that are weak acids (HIn) • HIn is one color while In is another color. • For example, phenolphthalein is colorless in the HIn form, but pink in the In form.

Applications of Aqueous Equilibria • HIn (Red) --> In (blue) + H+ • Ka = [H+][In-]/[HIn] • Rearrange to get Ka = [In-] [H+] [HIn] • Let’s say the indicator’s Ka = 1 x 10-8 • Add a few drops of the indicator to an acidic solution of pH = 1.0

Applications of Aqueous Equilibria 1 x 10-8 = [In-] 1 x 10-1 [HIn] 1 x 10-7 = 1 = [In-] 10,000,000 [HIn] • So in an acidic solution, most of the Indicator will be in the red HIn form. • As OH- is added, the [H+] dcreases, shifting the equilibrium so more In- and less HIn is present. • When will the color change, or rather, when can the human eye detect the color change? • At a pH when [In-] /[HIn] = 1/10

Applications of Aqueous Equilibria

Applications of Aqueous Equilibria