Chapter 15 Applications of Aqueous Equilibria • Some of what we do in this chapter will be the same as what we did in Chapter 14. • Let’s start with common ions. What do the weak acid HF and the salt NaF have in common?
The common ion effect • When the salt with the anion of a weak acid is added to that acid: • it reverses the dissociation of the acid • Lowers the % dissociation of the acid • This same principle applies to salts with the cation of a weak base as well. • Calculations are the same as last chapter
15.2 Buffered solutions • When an acid-base solution contains a common ion, it’s called a buffered solution. • A buffered solution is one that resists a change in pH with the addition of hydroxide ions or protons. • Often buffered solutions contain a weak acid and it’s salt (HF and NaF) OR a weak base and it’s salt (NH3 and NH4Cl). • We can make a buffer of any pH by varying the concentrations of these solutions.
Finding the pH of a buffered solution • Calculate the pH of a 1L buffered solution of 0.20M acetic acid solution and 0.30M sodium acetate. Ka=1.8x10-5
pH changes in a buffered solution • We will use the solution from the last problem, but add 0.01M NaOH. Compare the pH change that occurs with the addition of the NaOH solid. • There are 2 major step to proceed with these types of problems!! • 1. The stoichiometry • 2. The equilibrium
The stoichiometry1L of 0.20M acetic acid solution and 0.30M sodium acetate (Ka=1.8x10-5 ) [H+]=1.2x10-5 Buffered with .01M NaOH • Use moles not molarity • We will assume that all of the NaOH will be consumed
Now for the equilibrium - ICE • Since there is 1L of solution
pH change = 4.96 – 4.92 (from previous problem) Change in pH = .04 with the addition of a buffer.
Think about what would have happened…. • If base was added to just water • That means the pH of 0.01M NaOH in water is 12 • Minus the pH of water (7) means a change in pH of 5. • Compare adding NaOH to a buffered solution vs. an unbuffered solution….5 vs 0.04…
Remember • Buffered solutions are simply solutions of weak acids and weak bases containing a common ion • When a strong acid is added to a buffered soltuion…You know how to do these problems. Just remember the two steps… • Stoichiometry first • ICE (equilibrium) second
How does the buffer work? • Solve the equilibrium expression for [H+] • This means the [H+] depends on the ratio of [HA]/[A-]…if you take the –log of both sides…(I won’t bore you with the math)… you get…
It has a name! • It’s called the Henderson-Hasselback equation! • It also works for an acid and its salt, like HNO2 and NaNO2 • Or a base and its salt, like NH3 and NH4Cl, but you must remember to convert Ka to Kb in the equation
An assumption • One assumption of this formula is that the initial concentrations and the equilibrium concentrations are equivalent. (<5% validity) • It is a pretty safe assumption since the initial concentrations of HA and A- are relatively large in a buffered system.
Let’s try • What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH (formic acid and potassium formate) Ka=1.8x10-4
Given base, salt, and Kb • Calculate the pH of 0.25M NH3 and 0.40M NH4Cl. (Kb = 1.8 x 10-5) • Don’t forget to find Ka FIRST!! And the ratio is base over acid..
15.3 Buffer Capacity • The pH of a buffered solution is determined by the ratio of [A-]/[HA]. • If the ratio doesn’t change much…then the pH won’t change much either • The more concentrated these two are, the more H+ and OH- the solution will be able to absorb. • Larger concentrations = bigger buffer capacity
Try this problem. • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following: Ka= 1.8x10-5 5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc • Calculate initial pH for each solution (henderson-hasselbalch)
Now find the change after adding 0.04mol of HCl to 1.0L of solution • There’s virtually NO change. Now look at solution B. We already know the initial pH.
Solution B is 1L of 0.050 M HAc and 0.050 M NaAc • Compared to 4.74 before the acid was added. Solution A contains much larger quantities of buffering components and has a larger buffering capacity than solution B
Larger concentrations = bigger buffer capacity • Here’s why
Buffer Capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is the most resistant to change and true only when [A-] = [HA] • Makes pH = pKa (since the log of 1=0)
10.5 Titrations and pH Curves • Titration is commonly adding a solution of known concentration until the substance being tested is consumed. • This is often noted by a color change and is called the equivalence point. • This is often observed in a graph of pH vsmL of titrant and is called a titration curve.
Strong acid with a strong base titration • Net ionic equation H+ + OH-H2O • To know the amount of H+ at any point in the titration, we must determine the amount of H+ remaining and divide by the total volume of the solution. • Let’s first consider a new unit for molarity that is smaller as most titrations usually deal with mL Millimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L
The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with 0.100 MNaOH Where all the H+ ions originally present, have reacted with all the OH- ions added
Things to note • You need to do the stoichiometry for each step • mL x Molarity = mmol • There is no equilibrium. Both the acid and base dissociate completely • Use [H+] or [OH-] to figure pH or pOH • The equivalence point is when you have an equal number of moles of H+ and OH- • In other words enough H+ is present to react exactly with the OH- • Before the equivalence point H+ is in excess • After the equivalence point OH- is in excess
The pH Curve for the Titration of 100.0 mL of 0.50 MNaOH with 1.0 M HCI
Strong base titrated with strong acid. • Very similar to strong acid with a strong base except before the equivalence point the OH- is in excess and H+ is in excess after the equivalence point.
The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 MNaOHweak acid – strong base At equivalence point (pH > 7): A- is basic
Titrating a weak acid with a strong base • Again, there is no equilibrium • Do the stoichiometry – the reaction of OH- with the weak acid is assumed to run to completion & the concentrations of the acid remaining & conjugate base are determined. • Determine the major species • Since HA is stronger than H2O it is the dominant equilibrium • Then do the equilibrium (Henderson-Hasselbach)
The pH Curve for the Titrations of 100.0mL of 0.050 M NH3 with 0.10 M HClweak base – strong acid At equivalence point (pH < 7): NH4+ is acidic
In a titration curve • Equivalence point is defined by the stoichiometry NOT by the pH. • Remember it is where the mol of H+ are equal to the moles of OH-
Summary • Strong acid and base just stoichiometry. • Weak acid with 0 ml of base - Ka • Weak acid before equivalence point • Stoichiometry first • Then Henderson-Hasselbach • Weak acid at equivalence point- Kb -Calculate concentration • Weak acid after equivalence - leftover strong base. -Calculate concentration
Summary • Weak base with 0 ml of acid - Kb • Weak base before equivalence point. • Stoichiometry first • Then Henderson-Hasselbach • Weak base at equivalence point Ka. -Calculate concentration • Weak base after equivalence – left over strong acid. -Calculate concentration
15.5 Acid-Base Indicators • Two common methods for monitoring the pH • 1. pH meter • 2. acid-base indicator (this is not a good choice for finding the equivalence point.) Careful selection of indicator can help get results close. Endpoint IS change in color Equivalence point IS moles acid = moles base
HIn • Most indicators are themselves a weak acid. They are one color with the proton attached and another without the proton. • Phenolphthalein, the most common indicator, is colorless as an acid and pink as a base (In-)
Did the color change? • Typically, 1/10 of the initial form must be converted for the human eye to see a new color. When titrating an acidic solution…
If a basic solution is titrated • Indicator will initially exist as In- and more HIn will form when acid is added
Which indicator to use? • It is best to choose an indicator whose endpoint is closest to our equivalence point. • It is easier to choose an indictor if there is a large change in pH near the equivalence point (vertical area of pH curve) • The weaker the acid, the smaller the vertical area around the equivalence point, giving less flexibility in indicator choice
15.6 Solubility Equilibria and Ksp • Will it dissolve, and if not, how much? • If everything dissolves, it is an equilibrium position • Partial dissolving -- The solid will precipitate as fast as it dissolves (equilibrium) • Surface area changes the rate at which something dissolves, but NOT the amount (no change in equilibrium position)
Generic equation • M+ is the cation (usually metal) • Nm- is the anion (a nonmetal) • The solubility product for each compound is found below
Solubility vs. solubility product • Solubility is NOT the same as solubility product. • Solubility product is an equilibrium constant. • It doesn’t change except with temperature. • Solubility is an equilibrium position for how much can dissolve. • A common ion can change this.
Let’s try a couple…first an easy one! • What is the Ksp value of copper (I) bromide with a measured solubility of 2.0 x 10-4mol/L.
Now a bit more difficult • Calculate the Ksp for bismuth sulfide, which has a solubility of 1.0x10-15M at 25oC
Solubility from Ksp • Find the solubility of Silver Chloride. Ksp=1.6x10-10.
One more…a bit tougher • Find the solubility of copper (II) iodate, Cu(IO3)2. The Ksp for Cu(IO3)2 = 1.4x10-7 The solubility of copper (II) iodate is 3.3x10-3
Comparing Ksp values • Ksp values give info about solubility (relative solubility). • When salts have the SAME number of ions: the larger the Ksp the more soluble • When salts have different number of ions: you CANNOT compare directly.
Common ion effect and solubility • The presence of a common ion DECREASES the solubility of the salt. Let’s compare: What is the solubility of AgBr in both pure water & in 0.0010M NaBr. (pure H2O first)
Compare the two results In Water Common ion solution Notice that the common ion reduces the solubility as stated earlier (the reverse reaction occurs faster: equilibrium lies to the left)
remove add pH and solubility • Presence of a common ion decreases solubility Insoluble bases dissolve in acidic solutions • Insoluble acids dissolve in basic solutions At pH less than 10.45 Lower [OH-] Increase solubility of Mg(OH)2 At pH greater than 10.45 Raise [OH-] Decrease solubility of Mg(OH)2