🧪 Chemistry Exam 5 · Chapters 13–15
Chemistry · Exam 5

Exam 5 Study Guide

Exam 5 covers chapters 13, 14 and 15. All three chapters run on one idea. A reaction can settle part way and hold there. Once you can write the equilibrium constant and compare it with the reaction quotient, every problem in this unit opens up. Work through one chapter at a time. Do the worked examples with a pencil before you read the steps. Then take the self-test at the end of each chapter and check your arithmetic against the answers.

What this guide holds

  • Chapter 13: dynamic equilibrium, K and Q, Le Chatelier's principle, ICE tables.
  • Chapter 14: proton transfer, pH, strong and weak acids, Ka and Kb, salts, buffers, titration curves.
  • Chapter 15: Ksp and molar solubility, the common ion effect, precipitation, complex ions, amphoteric hydroxides, coupled equilibria.
  • Every equation with its units, three worked examples per chapter, the traps that cost the most points.
  • A 10-item self-test per chapter, and a one-page cram sheet at the end.

Chapter 13Fundamental Equilibrium Concepts

What to know

A reversible reaction runs both ways. It slows down and settles instead of finishing. At dynamic equilibrium the forward rate and the reverse rate are equal. The measured amounts stop changing. Molecules keep reacting the whole time. Nothing has stopped.

Equal rates do not mean equal amounts. One side can hold far more material than the other. Equilibrium also needs a closed container. If a product can leak away, the reverse rate never catches up. You reach the same resting mixture from either side, as long as the temperature is the same.

The equilibrium constant K puts a number on that resting state. Products go on top. Reactants go on the bottom. Each amount is raised to its own coefficient from the balanced equation. That rule is the law of mass action. K values carry no units.

Pure solids and pure liquids stay out of the expression. Their crowding cannot change as the reaction runs. Water used as the solvent drops out for the same reason. A reaction with more than one phase is a heterogeneous equilibrium.

The reaction quotient Q uses the same fraction as K. The difference is what you plug in. Q takes the amounts you have right now. Comparing Q with K gives you the direction of change, and there are only three outcomes.

The three-way test
  • Q < K: the top is too small, so the reaction runs forward and makes more product.
  • Q > K: the top is too large, so the reaction runs in reverse and rebuilds reactant.
  • Q = K: the mixture is already at rest, so nothing shifts.

Le Chatelier's principle is that same test in a hurry. Push on a system at equilibrium and it shifts the way that partly relieves the push. Add a species and the system moves away from it. Remove one and the system moves toward it. Squeezing a gas mixture shifts it toward the side with fewer moles of gas.

Temperature is the only stress that changes the value of K. Treat heat as a species in the equation. For an endothermic forward reaction, heat belongs on the left. Heating then shifts the reaction right and raises K. A catalyst speeds both directions by the same factor, so it never shifts the position. An unreactive gas added at fixed volume changes no concentration, so it does nothing.

An ICE table carries the bookkeeping for the math. The letters stand for Initial, Change and Equilibrium. You get one column per species and three rows. The Change entry is the coefficient times x, minus for reactants and plus for products. The Equilibrium row is the first two rows added. Feed that row into K and you have one equation in one unknown.

Sorting the four problem types
  • All the equilibrium amounts are given. Plug them in and report K.
  • Initial amounts plus one equilibrium amount are given. Find x from the known change, then fill the table.
  • K and the initial amounts are given. That is a full ICE table and some algebra.
  • Same as the one above, but K is tiny. Use the small-x approximation and then test it.

When K is very small, almost nothing reacts. The term (c0 − x) is nearly c0, so you can drop the x from that one spot. Only from that spot. The x values sitting alone are your answer, so they stay. Two tests guard the shortcut. Before you start, check that c0 ÷ K is above about 400. After you solve, use the five percent rule. If x is under five percent of c0, keep the answer.

Equations, units and symbols

EquationWhat each symbol meansUnits
Kc = [C]c[D]d ÷ ([A]a[B]b) Square brackets are molarity at equilibrium. Lower-case letters are the coefficients from the balanced equation. [ ] in mol/L. Kc has no units.
Kp = same fraction built from partial pressures Each gas enters as its own partial pressure instead of a concentration. Pressures in atm. Kp has no units.
Kp = Kc(RT)Δn Δn is moles of gas on the right minus moles of gas on the left. Count gases only. R = 0.08206 L·atm/mol·K, T in K, so RT is in L·atm/mol.
Q = same fraction as K, filled with the amounts present now Q is a snapshot. It drifts toward K and then stops there. No units.
Reverse the equation → K becomes 1 ÷ K Flipping the arrow inverts the fraction. No units.
Multiply the equation by n → K becomes Kn Scaling the coefficients raises K to that power. No units.
Add two equations → multiply their K values The shared species cancels, and the constants multiply. No units.
Equilibrium row = Initial row + Change row Change is the coefficient times x. Reactants lose, products gain. mol/L in every row.
Shortcut test: c0 ÷ K > 400 c0 is the starting concentration of the species that reacts. A bare ratio, no units.
Percent dissociation = (x ÷ c0) × 100% x is how much of the starting material reacted, in mol/L. A percent.
x = (−b ± √(b2 − 4ac)) ÷ 2a For ax2 + bx + c = 0. Keep only the root that leaves every concentration positive. x in mol/L.

Worked examples

Example 13.1 · Write K, convert it, then read Q

For 2 SO2(g) + O2(g) ⇌ 2 SO3(g), Kc = 2.80 × 102 at 1000 K. Write Kc, find Kp, then say which way a flask holding [SO2] = 0.100 M, [O2] = 0.200 M and [SO3] = 0.400 M will move.

Step 1 · write the fraction

Kc = [SO3]2 ÷ ([SO2]2[O2]). Every coefficient became an exponent.

Step 2 · count the gas moles

Right side has 2 moles of gas. Left side has 2 + 1 = 3. So Δn = 2 − 3 = −1.

Step 3 · build RT

RT = (0.08206 L·atm/mol·K)(1000 K) = 82.06 L·atm/mol.

Step 4 · raise and divide

A negative exponent means divide. Kp = (2.80 × 102) ÷ 82.06 = 3.41.

Step 5 · now find Q

Q = (0.400)2 ÷ [(0.100)2(0.200)]. Top: 0.160. Bottom: (0.0100)(0.200) = 0.00200. So Q = 0.160 ÷ 0.00200 = 80.0.

Answer

Kp = 3.41, and Q = 80.0 against Kc = 280. Q is below K, so the reaction runs forward and makes more SO3. Note that Kp came out far below Kc here. A negative Δn always does that.

Example 13.2 · A tiny K, so the shortcut holds

Sulfuryl chloride breaks up a little: SO2Cl2(g) ⇌ SO2(g) + Cl2(g), with Kc = 2.99 × 10-7. You start with 0.0500 M SO2Cl2. Find every equilibrium concentration.

Step 1 · run the ratio test

c0 ÷ K = 0.0500 ÷ (2.99 × 10-7) = 1.67 × 105. That is far above 400, so the shortcut is safe.

Step 2 · the table

Initial: 0.0500, 0, 0. Change: −x, +x, +x. Equilibrium: 0.0500 − x, x, x.

Step 3 · drop the one x

Kc = x2 ÷ (0.0500 − x) becomes x2 ÷ 0.0500 = 2.99 × 10-7.

Step 4 · solve

x2 = (2.99 × 10-7)(0.0500) = 1.495 × 10-8. Take the square root: x = 1.22 × 10-4 M.

Step 5 · check the five percent rule

(1.22 × 10-4) ÷ 0.0500 × 100% = 0.24%. Nowhere near 5%, so the answer stands.

Answer

[SO2] = [Cl2] = 1.22 × 10-4 M, and [SO2Cl2] = 0.0500 M to three figures. Check it: (1.22 × 10-4)2 ÷ 0.0499 = (1.49 × 10-8) ÷ 0.0499 = 2.99 × 10-7. It matches Kc.

Example 13.3 · When the shortcut fails

For PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 1.80 at 523 K. You start with 0.500 M PCl5 and nothing else. Find all three concentrations.

Step 1 · the ratio test warns you

c0 ÷ K = 0.500 ÷ 1.80 = 0.28. That is far under 400, so expect the shortcut to fail.

Step 2 · see how badly it fails

Dropping the x gives x2 ÷ 0.500 = 1.80, so x2 = 0.900 and x = 0.949 M. That is larger than the 0.500 M you started with, which is impossible.

Step 3 · set up the real quadratic

x2 ÷ (0.500 − x) = 1.80. Cross multiply: x2 = 0.900 − 1.80x. Rearranged: x2 + 1.80x − 0.900 = 0.

Step 4 · use the formula

b2 − 4ac = (1.80)2 + 4(1)(0.900) = 3.24 + 3.60 = 6.84. Its square root is 2.615. Then x = (−1.80 + 2.615) ÷ 2 = 0.815 ÷ 2 = 0.408 M.

Step 5 · fill in the bottom row

[PCl3] = [Cl2] = 0.408 M. [PCl5] = 0.500 − 0.408 = 0.092 M.

Answer and check

Check: (0.408)2 ÷ 0.092 = 0.166 ÷ 0.092 = 1.80. It matches. The percent dissociation is 0.408 ÷ 0.500 × 100% = 82%, so most of the PCl5 broke up. No x could be dropped here.

The traps people fall into

Watch for these
  • Calling equilibrium dead. The rates are equal, not zero. Molecules keep reacting both ways.
  • Putting a solid or the solvent into K. Pure solids, pure liquids and solvent water all stay out.
  • Losing a coefficient. A coefficient of 2 becomes an exponent of 2 in K and a change of 2x in the table.
  • Counting the wrong moles for Δn. Only gases count. If the gas moles match, Δn = 0 and Kp = Kc.
  • Using Celsius in RT. T must be in kelvin.
  • Expecting argon to do something. An unreactive gas added at fixed volume changes no concentration, so Q holds and nothing shifts.
  • Thinking a catalyst raises the yield. It shortens the wait and leaves K and the final amounts alone.
  • Dropping the x that is your answer. Drop x only where it is subtracted from a much larger number.
  • Filling the table with moles. Divide by the volume first, or your K will be wrong.
  • Keeping a root that gives a negative concentration. Reject it, and always put your answers back into K.

Self-test · Chapter 13

1. Write Kc for 2 NH3(g) ⇌ N2(g) + 3 H2(g).
Kc = [N2][H2]3 ÷ [NH3]2. Products on top, each raised to its coefficient. The 3 on hydrogen and the 2 on ammonia both come straight from the equation.
2. Write Kc and Kp for CaCO3(s) ⇌ CaO(s) + CO2(g).
Kc = [CO2] and Kp = PCO2. Both solids drop out, so one gas is all that is left. This also means a kiln reaches the same CO2 pressure no matter how much limestone you load.
3. For N2(g) + 3 H2(g) ⇌ 2 NH3(g), Kc = 0.0100 at 500 K. Find Kp.
Kp = 5.94 × 10-6. Gas moles: 2 on the right, 1 + 3 = 4 on the left, so Δn = −2. RT = (0.08206)(500) = 41.03 L·atm/mol, and (41.03)2 = 1683. Then Kp = 0.0100 ÷ 1683 = 5.94 × 10-6.
4. For A ⇌ B, K = 4.0. What is K for B ⇌ A, and for 2 A ⇌ 2 B?
0.25 and 16. Flipping gives 1 ÷ 4.0 = 0.25. Doubling the equation gives (4.0)2 = 16. The reaction did not change, only the way you wrote it.
5. H2(g) + I2(g) ⇌ 2 HI(g) has K = 50.0. A flask holds [H2] = [I2] = 0.100 M and [HI] = 1.00 M. Which way does it go?
In reverse. Q = (1.00)2 ÷ [(0.100)(0.100)] = 1.00 ÷ 0.0100 = 100. Since Q = 100 is above K = 50.0, there is too much product, so HI breaks back apart until Q falls to 50.0.
6. A flask at equilibrium holds [SO2] = 0.100 M, [O2] = 0.200 M and [SO3] = 0.400 M for 2 SO2(g) + O2(g) ⇌ 2 SO3(g). Find Kc.
Kc = 80.0. Kc = (0.400)2 ÷ [(0.100)2(0.200)]. Top: 0.160. Bottom: (0.0100)(0.200) = 0.00200. So 0.160 ÷ 0.00200 = 80.0. This is the easiest problem type, because the amounts given are already the equilibrium ones.
7. N2O4(g) ⇌ 2 NO2(g) takes in heat. A sealed tube is moved from ice water to hot water. What happens to the color and to K?
The brown deepens and K rises. Put heat on the left with the reactant. Heating adds to the left, so the reaction shifts right and makes more brown NO2. Temperature is the only stress that moves the value of K itself.
8. The piston on a flask of N2(g) + 3 H2(g) ⇌ 2 NH3(g) is pushed in to half the volume. Which way does the system shift?
To the right, toward ammonia. Squeezing shifts a gas reaction toward the side with fewer moles of gas. Here that is 2 moles on the right against 4 on the left. Factories use high pressure for exactly this reason.
9. COCl2(g) ⇌ CO(g) + Cl2(g) has Kc = 2.2 × 10-10. Start with 0.200 M COCl2 and find [CO].
[CO] = 6.6 × 10-6 M. Ratio test: 0.200 ÷ (2.2 × 10-10) = 9.1 × 108, so the shortcut is safe. Then x2 ÷ 0.200 = 2.2 × 10-10, so x2 = 4.4 × 10-11 and x = 6.6 × 10-6 M. That is 0.0033% of 0.200, well under 5%.
10. Start with 0.0100 M N2O4, where Kc = 4.6 × 10-3 for N2O4(g) ⇌ 2 NO2(g). Try the shortcut and judge it.
Reject it and use the quadratic. The shortcut gives (2x)2 ÷ 0.0100 = 4.6 × 10-3, so 4x2 = 4.6 × 10-5 and x = 3.4 × 10-3 M. That is 34% of 0.0100, far over the 5% limit. The ratio test said so first: 0.0100 ÷ (4.6 × 10-3) = 2.2, which is nowhere near 400.

Chapter 14Acid-Base Equilibria

What to know

A Bronsted-Lowry acid gives a proton away. A Bronsted-Lowry base takes one in. A proton here means H+, a hydrogen atom with its electron stripped off. Protons never stay loose in water. A water molecule grabs each one at once, making the hydronium ion, H3O+.

Give a proton away and you become the conjugate base of that acid. Take one in and you become the conjugate acid of that base. The two halves of a conjugate pair differ by one H+ and one unit of charge, and by nothing else. Every proton transfer holds two pairs, never one.

Water hands protons to itself now and then. That slow self-reaction is the autoionization of water. Its constant has its own name, Kw, and it equals 1.0 × 10-14 at 25 °C. The rule holds in every water solution, not just pure water. Push the hydronium up and the hydroxide must fall. Neither ion can ever reach zero.

Those exponents are awkward to say, so chemists take the negative base-ten logarithm. Pure water lands at pH 7.00. Below 7 a solution is acidic. Above 7 it is basic. Each whole pH unit is a tenfold change in hydronium, so pH 3 holds a hundred times more than pH 5.

Strong, weak, and the difference from concentration
  • A strong acid or base ionizes fully. The arrow points one way. Take the concentration straight to the pH.
  • The strong acids to know: HCl, HBr, HI, HNO3, HClO4, HClO3, and the first proton of H2SO4.
  • The strong bases to know: LiOH, NaOH and KOH give one OH- each. Ca(OH)2, Sr(OH)2 and Ba(OH)2 give two.
  • A weak acid or base reaches an equilibrium. Most of what you added is still whole.
  • Strength is a fraction. Concentration is an amount. A dilute strong acid can sit at a higher pH than a concentrated weak one.

Ka is the acid ionization constant, the equilibrium constant for a weak acid giving its proton to water. A larger Ka means a stronger acid. Bases get the matching Kb. Logs make the numbers easier to compare, so pKa = −log Ka, and a small pKa means a strong acid.

An acid and its conjugate base are locked together. Multiply the Ka of an acid by the Kb of its conjugate base and you always get Kw. So one table covers both, and a weaker acid always has a stronger conjugate base.

Percent ionization is the ionized share times 100. It is not a fixed property of the acid. It climbs as you dilute, because the ions get more room. The pH still rises, since there is less acid in total.

Polyprotic acids hold more than one donatable proton. They never let go of all of them at once. Each step has its own Ka, and each step is far harder than the one before. Pulling a positive proton off an already negative particle costs much more energy. That gap is a gift. Ka1 sets the pH almost by itself.

Dissolve a salt and its ions can react with water. That reaction is salt hydrolysis. An ion left over from a strong acid or a strong base is a spectator. An ion left over from a weak partner is not.

Reading a salt solution
  • Strong acid with strong base, such as NaCl: neutral.
  • Weak acid with strong base, such as sodium acetate: basic, because the anion takes protons.
  • Strong acid with weak base, such as NH4Cl: acidic, because the cation gives protons.
  • Weak with weak: compare the Ka of the cation with the Kb of the anion and let the larger one win.
  • A small metal ion with a high charge, such as Al3+: acidic, because its bound water gives up a proton.

A buffer mixes a weak acid with its conjugate base, both in real amounts. Add acid and the base half soaks up the protons. Add base and the acid half replaces them. Either way the pH barely moves. The Henderson-Hasselbalch equation is the fast route to a buffer pH. A buffer works best within about one pH unit of its pKa.

A titration curve plots pH against the volume of titrant added. The equivalence point is where the added titrant exactly matches what you started with. A strong acid against a strong base gives a near vertical jump through pH 7.00. A weak acid curve starts higher, has a long gentle buffer stretch, and lands above pH 7, because the salt left behind is a base. The most useful spot is the half-equivalence point, where the acid and its conjugate base are equal and pH = pKa.

Equations, units and symbols

EquationWhat each symbol meansUnits
Kw = [H3O+][OH-] = 1.0 × 10-14 The ion-product constant for water. Holds in any water solution at 25 °C. Each [ ] in mol/L. Kw has no units.
pH = −log[H3O+] The negative base-ten log of the hydronium concentration. pH is a bare number. [H3O+] in mol/L.
pOH = −log[OH-] The same move applied to hydroxide. pOH is a bare number.
pH + pOH = 14.00 Comes from taking the log of Kw. Good at 25 °C only. Bare numbers.
[H3O+] = 10-pH The trip back from a meter reading to a concentration. mol/L.
Ka = [H3O+][A-] ÷ [HA] HA is the intact weak acid. A- is its conjugate base. Liquid water is left out. [ ] in mol/L. Ka has no units.
Kb = [BH+][OH-] ÷ [B] B is the weak base. BH+ is its conjugate acid. No units.
Ka × Kb = Kw True for any conjugate pair, so pKa + pKb = 14.00 at 25 °C. No units.
pKa = −log Ka A small pKa means a strong acid. A bare number.
[H3O+] = √(Kac0) The shortcut for a weak acid, valid while x stays under 5% of c0. mol/L.
[OH-] = √(Kbc0) The same shortcut for a weak base. mol/L.
Percent ionization = ([H3O+] ÷ c0) × 100% c0 is the acid you poured in, before any reaction. A percent.
pH = pKa + log([A-] ÷ [HA]) The Henderson-Hasselbalch equation. Moles work as well as concentrations, since both share one volume. Bare numbers.
MaVa = MbVb Equivalence for a titration that trades one proton for one hydroxide. M in mol/L, V in L or mL as long as both match.

Worked examples

Example 14.1 · Both ends of the scale

Find the pH, the pOH and both ion concentrations for 0.0400 M HCl. Then do the same for 0.0400 M Ca(OH)2. Work at 25 °C.

Step 1 · the acid ionizes fully

HCl is strong, and each unit gives one proton. So [H3O+] = 0.0400 mol/L.

Step 2 · take the log

pH = −log(0.0400) = 1.40. Two digits after the point match the two significant figures in 0.0400.

Step 3 · finish the acid

pOH = 14.00 − 1.40 = 12.60. And [OH-] = (1.0 × 10-14) ÷ 0.0400 = 2.5 × 10-13 mol/L.

Step 4 · count hydroxides per formula unit

Each Ca(OH)2 unit sets two hydroxides loose. So [OH-] = 2 × 0.0400 = 0.0800 mol/L.

Step 5 · pOH, then pH

pOH = −log(0.0800) = 1.10, so pH = 14.00 − 1.10 = 12.90.

Answer and check

The acid sits at pH 1.40 and the base at pH 12.90. Multiply the acid's two ion amounts: (0.0400)(2.5 × 10-13) = 1.0 × 10-14, which is Kw. Skipping the factor of two on the base would have given pH 12.60, wrong by 0.30.

Example 14.2 · A weak acid, with the shortcut tested

Hydrofluoric acid has Ka = 6.8 × 10-4. Find the pH and the percent ionization of a 0.500 M solution.

Step 1 · write the equilibrium

HF + H2O ⇌ H3O+ + F-. Let x be the hydronium made, in mol/L.

Step 2 · the table

At balance, [H3O+] = x, [F-] = x, and [HF] = 0.500 − x.

Step 3 · try the shortcut

c0 ÷ Ka = 0.500 ÷ (6.8 × 10-4) = 735, just above 400. So x2 ÷ 0.500 = 6.8 × 10-4, giving x2 = 3.4 × 10-4 and x = 0.0184 M.

Step 4 · check the five percent rule

0.0184 ÷ 0.500 × 100% = 3.7%. Under 5%, so the shortcut is allowed.

Step 5 · pH and percent ionization

pH = −log(0.0184) = 1.73. The percent ionization is the same 3.7% from step 4.

Answer

pH 1.73 and 3.7% ionized. The full quadratic gives x = 0.0181 M and pH 1.74, so the shortcut cost one hundredth of a pH unit. Notice that 96% of the acid is still intact HF.

Example 14.3 · A buffer, before and after an attack

One litre holds 0.200 mol acetic acid and 0.300 mol sodium acetate. The pKa of acetic acid is 4.74. Find the pH. Then add 0.0500 mol HCl and find it again.

Step 1 · read the starting ratio

[A-] ÷ [HA] = 0.300 ÷ 0.200 = 1.50, and log(1.50) = 0.18.

Step 2 · the starting pH

pH = 4.74 + 0.18 = 4.92. The base half outnumbers the acid half, so the pH sits above pKa.

Step 3 · react the strong acid

Added protons turn acetate back into acetic acid, one for one. Acetate falls to 0.300 − 0.0500 = 0.250 mol. Acetic acid rises to 0.200 + 0.0500 = 0.250 mol.

Step 4 · the new pH

pH = 4.74 + log(0.250 ÷ 0.250) = 4.74 + log(1) = 4.74 + 0 = 4.74.

Answer and compare

The pH fell from 4.92 to 4.74, a move of 0.18 units. That same 0.0500 mol HCl in one litre of pure water would give [H3O+] = 0.0500 and pH 1.30, a drop of 5.70 units from neutral. The buffer is doing real work.

The traps people fall into

Watch for these
  • Forgetting the second hydroxide. Ba(OH)2, Ca(OH)2 and Sr(OH)2 each release two.
  • Treating a weak acid as strong. A 0.10 M acetic acid solution is pH 2.87, not 1.00.
  • Mixing strength with concentration. 0.0010 M HCl is pH 3.00 while 0.10 M acetic acid is pH 2.87. The weak one is lower here.
  • Counting significant figures wrong in a pH. Only the digits after the decimal point count, so pH 2.60 carries two.
  • Using Ka when the problem hands you the conjugate. Convert first with Ka × Kb = Kw.
  • Reaching for pH + pOH = 14.00 away from 25 °C. Kw changes with temperature, so that sum does too.
  • Calling every salt neutral. Check where each ion came from before you answer.
  • Using Ka2 for the pH of a polyprotic acid. Ka1 does nearly all the work.
  • Using Henderson-Hasselbalch where it does not apply. It needs both members of the pair present in real amounts.
  • Expecting a weak acid titration to end at pH 7. It lands above 7, because the salt left behind is a base.

Self-test · Chapter 14

1. A solution has [OH-] = 2.5 × 10-4 M at 25 °C. Find the pH.
pH = 10.40. pOH = −log(2.5 × 10-4) = 3.60. Then pH = 14.00 − 3.60 = 10.40. A pH above 7 fits a solution rich in hydroxide.
2. A meter reads pH 3.70. Find [H3O+] and [OH-].
2.0 × 10-4 M and 5.0 × 10-11 M. [H3O+] = 10-3.70 = 2.0 × 10-4 M. Then [OH-] = (1.0 × 10-14) ÷ (2.0 × 10-4) = 5.0 × 10-11 M. Their product is 1.0 × 10-14, which is Kw.
3. Find the pH of 0.0200 M Sr(OH)2.
pH = 12.60. Each unit gives two hydroxides, so [OH-] = 2(0.0200) = 0.0400 M. Then pOH = −log(0.0400) = 1.40, and pH = 14.00 − 1.40 = 12.60.
4. Hydrocyanic acid has Ka = 4.9 × 10-10. Find the pH of a 0.100 M solution.
pH = 5.15. The ratio 0.100 ÷ (4.9 × 10-10) = 2.0 × 108 clears 400 easily. Then x2 = (4.9 × 10-10)(0.100) = 4.9 × 10-11, so x = 7.0 × 10-6 M and pH = −log(7.0 × 10-6) = 5.15. Almost none of it ionized.
5. Ammonia has Kb = 1.8 × 10-5. Find Ka and pKa for the ammonium ion.
Ka = 5.6 × 10-10 and pKa = 9.25. Ka = Kw ÷ Kb = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10. Then pKa = −log(5.6 × 10-10) = 9.25. Check it the other way: Ka × Kb = (5.6 × 10-10)(1.8 × 10-5) = 1.0 × 10-14, which is Kw.
6. Find the pH of 0.10 M NaF. Hydrofluoric acid has Ka = 6.8 × 10-4.
pH = 8.08. Sodium is a spectator and fluoride is a base. Kb = (1.0 × 10-14) ÷ (6.8 × 10-4) = 1.47 × 10-11. Then x2 = (1.47 × 10-11)(0.10) = 1.47 × 10-12, so [OH-] = 1.2 × 10-6 M. pOH = 5.92, so pH = 8.08. A salt of a weak acid should come out basic, and it did.
7. Is a solution of NH4NO3 acidic, basic or neutral? Why?
Acidic. Nitrate came from nitric acid, a strong acid, so it is a spectator. Ammonium came from ammonia, a weak base, so ammonium is an acid and gives a proton to water. One active ion decides the answer.
8. 500 mL holds 0.100 mol formic acid and 0.200 mol sodium formate. Formic acid has pKa = 3.74. Find the pH.
pH = 4.04. pH = 3.74 + log(0.200 ÷ 0.100) = 3.74 + log(2.00) = 3.74 + 0.30 = 4.04. The volume never entered the math, because both members share it. The pH sits 0.30 units above the pKa, which is what a 2-to-1 ratio buys you.
9. 0.100 M acetic acid is titrated with NaOH. Acetic acid has pKa = 4.74. What is the pH at the half-equivalence point, and why?
pH = 4.74. Half the acid has become acetate, so [A-] = [HA]. The ratio is 1 and log(1) = 0, which leaves pH = pKa. This is why a titration curve is the easiest way to measure a pKa.
10. How many mL of 0.0500 M NaOH does it take to reach equivalence with 25.00 mL of 0.100 M HCl?
50.00 mL. Moles of acid = (0.02500 L)(0.100 mol/L) = 2.50 × 10-3 mol. The trade is one to one, so you need the same moles of base. Volume = (2.50 × 10-3 mol) ÷ (0.0500 mol/L) = 0.05000 L = 50.00 mL. Half the concentration means twice the volume.

Chapter 15Equilibria of Other Reaction Classes

What to know

Stir a stubborn salt into water and some of it sits on the bottom. That leftover solid is not resting. Ions keep leaving its surface and others keep landing back. When the two rates match, the solution is saturated. So a saturated solution is an equilibrium, and it has a constant.

That constant is the solubility product constant, Ksp. Build it the usual way. Multiply the ion concentrations on the product side, each raised to its coefficient. The solid never appears. A pure solid has fixed crowding, so a mountain of it and a speck of it give the same Ksp. Adding more solid cannot force more into solution.

Molar solubility s is the moles of a salt that dissolve in one liter to reach saturation. The bridge between s and Ksp is the formula of the salt, so write the balanced dissolving equation every time.

The shape of Ksp by ion ratio
  • 1:1, such as AgCl, gives Ksp = s2.
  • 1:2 or 2:1, such as PbI2 or Ag2CrO4, gives Ksp = 4s3.
  • 3:2, such as Ca3(PO4)2, gives Ksp = 108s5.
  • Compare two Ksp values head to head only when the salts share the same ion ratio.

Real water is rarely pure. An ion that two solutes both supply is a common ion. Raise the amount of a product and the dissolving equilibrium shifts back toward solid. That drop is the common ion effect. Ksp itself does not change. What changes is how the two concentrations split the product. When the shared ion is squared in Ksp, the squeeze is far stronger.

Pour two clear solutions together and a precipitate may appear. You can predict it. Build the same product that Ksp uses, but from the concentrations you actually have. That is the ion product Q. Below Ksp the solution is unsaturated and no solid forms. Above it the solution is supersaturated and solid drops out until Q falls back.

One step gets skipped a lot. Mixing two solutions dilutes both of them. Find each ion's new concentration in the combined volume before you build Q.

If two ions both form solids with the same reagent, they do not come out together. Add the reagent slowly and the less soluble salt forms first. That is selective precipitation. The method is to find the reagent concentration each salt needs to just begin. Whichever salt needs less goes first.

A complex ion is a central metal ion bonded to several small molecules or ions. Each attached partner is a ligand, and every ligand has a lone pair to donate. So the ligand is a Lewis base and the metal ion is a Lewis acid. The number of donor atoms attached is the coordination number. Silver likes 2, zinc and copper often take 4, and many first-row ions take 6.

Complex formation has its own constant, the formation constant Kf. These values are large, often huge. A big Kf means almost every metal ion ends up inside a complex, so the free ion concentration drops to nearly nothing. That is how ammonia keeps silver in solution.

Some hydroxides dissolve in acid and in strong base. Such a solid is amphoteric. With acid, the hydroxide ions get taken and the metal leaves as a simple cation. With extra base, the metal picks up one more hydroxide as a ligand and leaves as a soluble hydroxo complex. Plot dissolved metal against pH and you get a V. Aluminum, zinc, chromium(III), beryllium, tin(II) and lead(II) hydroxides do this. Magnesium and iron(III) hydroxides do not.

Two equilibria that share a species are coupled equilibria. The bookkeeping never changes. Add the two balanced equations, cancel the shared species, and multiply the two constants. Flip an equation and flip its constant to 1 ÷ K. Acid dissolves salts whose anion is the conjugate base of a weak acid, such as carbonate, phosphate, sulfide and hydroxide. Acid does nothing for chloride, bromide, iodide or nitrate salts. For those, go after the metal with a ligand instead.

Equations, units and symbols

EquationWhat each symbol meansUnits
Ksp = [Mn+]x[Xm-]y Only the dissolved ions appear, each raised to its coefficient. The solid is left out. [ ] in mol/L. Ksp has no units.
1:1 salt: Ksp = s2, so s = √Ksp s is molar solubility. Both ions sit at s. s in mol/L.
1:2 or 2:1 salt: Ksp = 4s3 One ion sits at s and the other at 2s, and 2s is squared. s in mol/L.
3:2 salt: Ksp = 108s5 From (3s)3(2s)2 = 27s3 × 4s2. s in mol/L.
Grams per liter = s × molar mass The mass that dissolves in one liter at saturation. g/L, from mol/L times g/mol.
Q = the same ion product, built from the amounts present now Q < Ksp: unsaturated. Q = Ksp: saturated. Q > Ksp: solid forms. No units.
Diluted concentration = original × (its volume ÷ total volume) Do this to every ion before building Q for a mixture. mol/L.
Kf = [MLn] ÷ ([M][L]n) M is the free metal ion, L is the free ligand, MLn is the complex. Kd = 1 ÷ Kf. [ ] in mol/L. Kf has no units.
Koverall = K1 × K2 Add two equations and multiply their constants. Flipping one gives 1 ÷ K. No units.

Worked examples

Example 15.1 · Ksp to solubility, and why Ksp is not a ranking

Silver chromate has Ksp = 9.0 × 10-12 at 25 °C, and its molar mass is 331.7 g/mol. Find its molar solubility and its solubility in g/L. Then compare it with AgCl, where Ksp = 1.8 × 10-10.

Step 1 · balance the dissolving

Ag2CrO4(s) ⇌ 2 Ag+(aq) + CrO42-(aq).

Step 2 · label the ions with s

Each formula unit gives two silvers and one chromate. So [Ag+] = 2s and [CrO42-] = s.

Step 3 · build Ksp

Ksp = (2s)2(s) = 4s3 = 9.0 × 10-12.

Step 4 · solve for s

s3 = (9.0 × 10-12) ÷ 4 = 2.25 × 10-12. The cube root gives s = 1.3 × 10-4 M.

Step 5 · convert to mass

(1.31 × 10-4 mol/L)(331.7 g/mol) = 0.043 g/L. Also [Ag+] = 2s = 2.6 × 10-4 M.

Answer and the comparison

Check it backward: 4(1.31 × 10-4)3 = 9.0 × 10-12. For AgCl, a 1:1 salt, s = √(1.8 × 10-10) = 1.3 × 10-5 M. So AgCl has the larger Ksp and the smaller solubility, by a factor of ten. The exponent is why. A cube pulls a small number down much harder than a square does.

Example 15.2 · A common ion shuts the door

Calcium fluoride has Ksp = 3.9 × 10-11. Find its molar solubility in pure water, then in 0.0500 M NaF.

Step 1 · pure water first

CaF2(s) ⇌ Ca2+ + 2 F-, so Ksp = 4s3. Then s3 = (3.9 × 10-11) ÷ 4 = 9.75 × 10-12, and s = 2.1 × 10-4 M.

Step 2 · set up the salty case

Let s be the moles of CaF2 that dissolve per liter. Then [Ca2+] = s and [F-] = 0.0500 + 2s.

Step 3 · drop the tiny term

In pure water s was only 2.1 × 10-4 M, and here it will be far smaller. So 0.0500 + 2s is near enough 0.0500.

Step 4 · solve

Ksp = (s)(0.0500)2 = 0.00250s = 3.9 × 10-11. So s = (3.9 × 10-11) ÷ 0.00250 = 1.6 × 10-8 M.

Step 5 · test the shortcut

2s = 3.2 × 10-8 M, which is nothing next to 0.0500 M. The shortcut was fair.

Answer

1.6 × 10-8 M in the fluoride solution, against 2.1 × 10-4 M in pure water. The shared ion cut solubility by about 13,000 times. Fluoride is squared in Ksp, which is why the cut is so deep.

Example 15.3 · Will a solid drop out?

You mix 100.0 mL of 1.0 × 10-4 M BaCl2 with 100.0 mL of 1.0 × 10-4 M Na2SO4. Does BaSO4 form? Its Ksp is 1.1 × 10-10.

Step 1 · total volume

100.0 mL + 100.0 mL = 200.0 mL.

Step 2 · dilute each ion

Each solution lost half its strength. [Ba2+] = (1.0 × 10-4)(100.0 ÷ 200.0) = 5.0 × 10-5 M. The sulfate works out the same way, at 5.0 × 10-5 M.

Step 3 · build Q

Q = [Ba2+][SO42-] = (5.0 × 10-5)(5.0 × 10-5) = 2.5 × 10-9.

Step 4 · compare

Q = 2.5 × 10-9 against Ksp = 1.1 × 10-10. The ratio is 2.5 ÷ 0.11 = 23, so Q is about 23 times Ksp.

Answer

Q is above Ksp, so white BaSO4 precipitates. It keeps forming until Q falls back to 1.1 × 10-10. Skipping the dilution step would have given Q = 1.0 × 10-8, four times too large.

The traps people fall into

Watch for these
  • Writing the solid into Ksp. It is never there, so piling in more solid cannot dissolve more.
  • Ranking salts by Ksp across different ion ratios. Ag2CrO4 has the smaller Ksp and the larger solubility.
  • Forgetting the 4 in 4s3. The 2s gets squared, which brings a factor of 4 along.
  • Skipping the dilution. Mixing two solutions weakens both before anything reacts.
  • Using the added ion without checking s. The shortcut only holds while s is far smaller than the amount already there.
  • Expecting acid to dissolve any salt. Chloride, bromide, iodide and nitrate salts hold no protons at ordinary pH.
  • Treating hydroxide as only a common ion. A trace of it lowers solubility, and a flood of it raises solubility again as a ligand.
  • Flipping an equation and keeping its constant. Flip the equation and the constant becomes 1 ÷ K.
  • Ignoring the ligand used up by the complex. Each complex ties up several ligands, so subtract them before you plug in.
  • Reading a V-shaped curve backward. The floor of the V is the pH where the metal is hardest to dissolve.

Self-test · Chapter 15

1. Write Ksp for Ca3(PO4)2, then write it in terms of s.
Ksp = [Ca2+]3[PO43-]2 = 108s5. Each unit gives three calcium ions at 3s and two phosphates at 2s. So (3s)3(2s)2 = 27s3 × 4s2 = 108s5.
2. Silver iodide has Ksp = 8.3 × 10-17. Find its molar solubility.
9.1 × 10-9 M. AgI is 1:1, so Ksp = s2 and s = √(8.3 × 10-17). Rewrite that as √(83 × 10-18) = 9.1 × 10-9 M. Shifting the exponent to an even power makes the root easy to take.
3. Lead(II) chloride has Ksp = 1.7 × 10-5. Find its molar solubility.
1.6 × 10-2 M. PbCl2 is 1:2, so Ksp = 4s3. Then s3 = (1.7 × 10-5) ÷ 4 = 4.25 × 10-6, and the cube root is 1.6 × 10-2 M. Also [Cl-] = 2s = 3.2 × 10-2 M.
4. Magnesium hydroxide has Ksp = 5.6 × 10-12. Find the pH of a saturated solution.
pH = 10.35. Mg(OH)2 is 1:2, so 4s3 = 5.6 × 10-12 and s3 = 1.4 × 10-12, giving s = 1.1 × 10-4 M. Then [OH-] = 2s = 2.2 × 10-4 M, so pOH = 3.65 and pH = 14.00 − 3.65 = 10.35. This is why milk of magnesia is mildly basic.
5. Barium sulfate has Ksp = 1.1 × 10-10. Find its solubility in 0.0100 M Na2SO4, and compare with pure water.
1.1 × 10-8 M, down from 1.0 × 10-5 M. In the salt solution, (s)(0.0100) = 1.1 × 10-10, so s = 1.1 × 10-8 M. In pure water s = √(1.1 × 10-10) = 1.0 × 10-5 M. The shared sulfate cut solubility by about 950 times.
6. Equal volumes of 2.0 × 10-3 M AgNO3 and 2.0 × 10-3 M NaCl are mixed. Does AgCl form? Ksp = 1.8 × 10-10.
Yes. Equal volumes halve each concentration, so [Ag+] = [Cl-] = 1.0 × 10-3 M. Then Q = (1.0 × 10-3)2 = 1.0 × 10-6. Since Q is about 5,600 times Ksp, a heavy white precipitate forms.
7. A solution is 0.010 M in Cl- and 0.010 M in CrO42-. Silver nitrate is added drop by drop. Which solid forms first? Ksp(AgCl) = 1.8 × 10-10 and Ksp(Ag2CrO4) = 9.0 × 10-12.
AgCl forms first. AgCl needs [Ag+] = (1.8 × 10-10) ÷ 0.010 = 1.8 × 10-8 M. Ag2CrO4 needs [Ag+] = √[(9.0 × 10-12) ÷ 0.010] = √(9.0 × 10-10) = 3.0 × 10-5 M. AgCl needs far less silver, so it goes first. Note that the smaller Ksp did not win, because the ion ratios differ.
8. A solution holds 1.0 × 10-3 M total silver and 0.100 M NH3. Find the free [Ag+]. Kf for [Ag(NH3)2]+ is 1.7 × 107.
6.1 × 10-9 M. Kf is huge and ammonia is in excess, so treat almost all the silver as complexed. Each complex uses 2 NH3, so free [NH3] = 0.100 − 0.0020 = 0.098 M. Rearranged, [Ag+] = (1.0 × 10-3) ÷ [(1.7 × 107)(0.098)2]. The bracket is (1.7 × 107)(0.0096) = 1.6 × 105, so [Ag+] = 6.1 × 10-9 M.
9. Find K for Mg(OH)2(s) + 2 H3O+ ⇌ Mg2+ + 4 H2O. Use Ksp = 5.6 × 10-12 and Kw = 1.0 × 10-14.
K = 5.6 × 1016. Step one is the dissolving, with Ksp = 5.6 × 10-12. Step two is H3O+ + OH- ⇌ 2 H2O, which is the reverse of autoionization, so its constant is 1 ÷ Kw = 1.0 × 1014. You need that step twice, so square it: 1.0 × 1028. Multiply: (5.6 × 10-12)(1.0 × 1028) = 5.6 × 1016. A K that large means acid clears the solid away at once.
10. Aluminum hydroxide is amphoteric. What forms when you add acid, and what forms when you add excess NaOH?
Acid gives Al3+, and excess base gives [Al(OH)4]-. With acid, hydronium takes the hydroxide from the solid and the metal leaves as a plain cation. With a flood of hydroxide, one more OH- attaches as a ligand and the metal leaves inside a soluble hydroxo complex. Plot the dissolved metal against pH and you get a V, with the floor near pH 6 for aluminum.

Cram sheet

One page for the last pass before the exam. Every constant below is for 25 °C unless the line says otherwise.

Constants worth knowing cold

  • R = 0.08206 L·atm/mol·K, and T must be in kelvin, where 0 °C = 273.15 K.
  • Kw = [H3O+][OH-] = 1.0 × 10-14, so pH + pOH = 14.00.
  • Pure water: [H3O+] = [OH-] = 1.0 × 10-7 mol/L, pH 7.00.
  • The shortcut is safe when c0 ÷ K is above about 400, and x must stay under 5% of c0.
  • Acetic acid: Ka = 1.8 × 10-5, pKa 4.74. Ammonia: Kb = 1.8 × 10-5, so ammonium has Ka = 5.6 × 10-10.

Every equation on one screen

Use it forEquation
Equilibrium constantKc = products over reactants, each raised to its coefficient
Gas constantsKp = Kc(RT)Δn, Δn = gas moles right minus left
Direction of changeQ < K forward, Q > K reverse, Q = K at rest
Rewriting an equationreverse → 1 ÷ K; times n → Kn; add → multiply
Solving for xx = (−b ± √(b2 − 4ac)) ÷ 2a
How far it wentpercent dissociation = (x ÷ c0) × 100%
Water's two ionsKw = [H3O+][OH-] = 1.0 × 10-14
The pH scalepH = −log[H3O+], [H3O+] = 10-pH, pH + pOH = 14.00
Weak acid or baseKa = [H3O+][A-] ÷ [HA], Kb = [BH+][OH-] ÷ [B]
Fast weak-acid pH[H3O+] = √(Kac0), and [OH-] = √(Kbc0)
Conjugate pairsKa × Kb = Kw, pKa + pKb = 14.00
BufferspH = pKa + log([A-] ÷ [HA]); pH = pKa at half equivalence
TitrationMaVa = MbVb for a one-to-one trade
SolubilityKsp = s2 (1:1), 4s3 (1:2 or 2:1), 108s5 (3:2)
Precipitationbuild Q from diluted amounts, then compare it with Ksp
Complex ionsKf = [MLn] ÷ ([M][L]n), Kd = 1 ÷ Kf
Coupled systemsadd the equations, multiply the constants

Five moves that solve most problems

  • Write the balanced equation first. Every exponent and every change entry comes from it.
  • Find Q before you set up a table. It tells you the sign of x.
  • Run the ratio test, solve, then check the five percent rule.
  • Convert between Ka and Kb with Kw whenever a problem hands you the conjugate.
  • Put the answer back into K. If it does not reproduce K, something is wrong.

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