🧪 Contents · Interactive Lab Chapter 14 · Acid-Base Equilibria
Chemistry · Chapter 14

Acid-Base Equilibria

Lemon juice, soap, blood and soda all sit on one scale. This chapter shows what that scale measures. The whole story is one proton. Acids hand a proton over. Bases take one in. Water does both, which is why water sets the middle of the scale. Once you can follow that single proton, pH stops being a mystery number. You will find pH from a concentration, and a concentration from pH. You will see why some acids give up every proton and others barely let go. By the end you can mix a buffer and read a titration curve.

By the end of this chapter you can

  • Point to the acid, the base and both conjugate pairs in any proton transfer.
  • Use Kw to link [H3O+], [OH-], pH and pOH at 25 °C.
  • Tell a strong acid or base from a weak one, and say why the difference shows up in pH.
  • Find pH and percent ionization from Ka or Kb and a starting concentration.
  • Swap between Ka and Kb using Kw, and between pKa and pKb.
  • Predict whether a salt solution comes out acidic, basic or neutral.
  • Mix a buffer with the Henderson-Hasselbalch equation and read any titration curve.

Section 1Bronsted-Lowry acids, bases and conjugate pairs

One proton changes hands

A Bronsted-Lowry acid is a particle that gives a proton away. A Bronsted-Lowry base is a particle that takes one in. That is the whole rule. A proton here means H+. It is a hydrogen atom with its one electron stripped off. What is left is tiny and carries a full positive charge.

Protons never drift around loose in water. The instant one breaks free, a nearby water molecule grabs it. The result is the hydronium ion, H3O+. So when a book writes H+ in water, picture H3O+. Hydrogen chloride shows the handoff plainly.

HCl + H2O → Cl- + H3O+

HCl gave the proton away, so HCl is the acid. Water took it in, so water is the base. Nothing else moved. One proton crossed, and both partners got a new name.

Every acid leaves a base behind

Give up a proton and you turn into something new. That leftover is the conjugate base of the acid. Take a proton in and you become the conjugate acid of the base. The two halves of a conjugate acid-base pair differ by one H+. They differ in nothing else.

So every proton transfer holds two pairs. Watch hydrofluoric acid in water.

HF + H2O ⇌ F- + H3O+

HF and F- are one pair. Water and H3O+ are the other. To find a partner, take off one hydrogen and one plus of charge. HF minus H+ leaves F-. That check works every time.

The double arrow matters here. HF lets go of only a few of its protons. Fluoride takes some of them right back. The mixture settles into a balance, so you write ⇌ and not a single arrow.

One proton moves and makes two conjugate pairs one H⁺ crosses over HF + H2O ⇌ F− + H3O+ acid base conjugate base conjugate acid pair 1: HF / F− pair 2: H₂O / H₃O⁺ Each pair differs by one hydrogen and one unit of charge.
Figure 14.1 One proton leaves HF and lands on water. That single move creates two conjugate pairs at once. Subtract one H and one plus of charge to find a partner.

Water plays both parts

Water donated nothing in the HF reaction. It only accepted. Yet water hands a proton to ammonia without any fuss, making NH4+ and OH-. A species that can both give and take a proton is amphiprotic. Water is the one you will meet most.

Plenty of ions behave the same way. Bicarbonate, HCO3-, is a clear case. It can give its proton to water and become CO32-. It can also take a proton from water and become H2CO3. Dihydrogen phosphate, H2PO4-, does both too. Which path wins depends on two numbers you will meet in section 5.

Worked example · label both pairs

Bicarbonate acts as a base in this reaction. Name the acid, the base, and both conjugate pairs.

HCO3- + H2O ⇌ H2CO3 + OH-

Step 1 · track the hydrogenCompare each left side particle with its right side partner. HCO3- gained a hydrogen to become H2CO3. Water lost one to become OH-.

Step 2 · name the rolesWater gave the proton away, so water is the acid here. Bicarbonate took it in, so bicarbonate is the base.

Step 3 · write the pairsThe pairs are H2CO3 with HCO3-, and H2O with OH-. Each pair differs by one H+.

CheckCharge must balance. Left side is 1 minus. Right side is 0 plus 1 minus. Both sides match.

Key points
  • An acid donates a proton. A base accepts one. Water can do either job.
  • A conjugate pair differs by exactly one H+ and one unit of charge.
  • Every proton transfer contains two conjugate pairs, never one.
  • A loose proton in water is really a hydronium ion, H3O+.
Interactive Pass a proton and label both conjugate pairs. Open full ↑

Section 2Water autoionization and the pH scale

Water splits itself

Pure water is not just H2O molecules. Now and then one water molecule hands a proton to another. One becomes H3O+ and the other becomes OH-. This slow self-reaction is called the autoionization of water.

2 H2O ⇌ H3O+ + OH-

Its equilibrium constant gets its own name and symbol. The ion-product constant for water is Kw, and it equals the product of the two ion concentrations.

Kw = [H3O+][OH-] = 1.0 × 10-14 at 25 °C

In pure water the two ions come from the same split. So they must be equal. Each one works out to 1.0 × 10-7 mol/L, because that number squared gives 1.0 × 10-14. A solution with equal amounts of the two ions is neutral.

Here is the useful part. Kw holds in every water solution, not just pure water. Push the hydronium up and the hydroxide must fall. The product stays fixed. One ion can never go to zero.

pH shrinks the numbers

Those exponents are awkward to say out loud. So chemists take the negative base-ten logarithm. pH is −log[H3O+], and pOH is −log[OH-]. Taking the log of Kw gives a tidy partner rule.

pH + pOH = 14.00 at 25 °C

Pure water lands at pH 7.00 and pOH 7.00. Below pH 7 a solution is acidic. Above pH 7 it is basic. Each whole pH unit is a tenfold change in hydronium. So pH 3 holds a hundred times more hydronium than pH 5.

The pH scale with matching hydronium and hydroxide amounts 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH neutral 7.00 more acidic more basic [H₃O⁺] mol/L 100 10−2 10−4 10−6 10−8 10−10 10−12 10−14 [OH−] mol/L 10−14 10−12 10−10 10−8 10−6 10−4 10−2 100 stomach acid 1.0 lemon juice 2.4 cola 3.3 black coffee 5.0 blood 7.4 baking soda 8.4 ammonia spray 11.0 bleach 12.6 Multiply the two rows and you always get 1.0 × 10−¹⁴ at 25 °C.
Figure 14.2 One step along pH is a tenfold change in hydronium. The two concentration rows move in opposite directions. Their product never changes at a fixed temperature.

Worked example · pH of dilute nitric acid

You have 0.0025 M HNO3. Find [H3O+], pH, pOH and [OH-] at 25 °C.

Step 1 · hydroniumNitric acid is strong, so it ionizes fully. Each HNO3 unit gives one proton. So [H3O+] = 2.5 × 10-3 mol/L.

Step 2 · pHpH = −log(2.5 × 10-3) = 2.60. Two digits after the point match the two significant figures in 2.5.

Step 3 · pOHpOH = 14.00 − 2.60 = 11.40.

Step 4 · hydroxide[OH-] = Kw ÷ [H3O+] = (1.0 × 10-14) ÷ (2.5 × 10-3) = 4.0 × 10-12 mol/L.

CheckMultiply the two ion amounts. (2.5 × 10-3)(4.0 × 10-12) = 1.0 × 10-14. That is Kw, so the answer holds.

The trip also runs backward. Suppose a cleaner reads pH 10.50 on a meter. Raise 10 to the power of the negative pH, and [H3O+] = 10-10.50 = 3.2 × 10-11 mol/L. The pOH is then 3.50, which puts [OH-] at 3.2 × 10-4 mol/L.

Why a log scale is worth the trouble

Real solutions cover a huge range of hydronium amounts. Stomach acid holds a few million times more hydronium than blood does. Writing those two numbers side by side takes a pile of zeros, and comparing them by eye is hopeless. The log scale folds that whole range into the familiar 0 to 14 stretch, so a difference of two units always means the same thing: a factor of a hundred.

Keep one habit from this. The digits after the decimal point in a pH are the only significant figures it has. A pH of 2.60 carries two of them, which is why it came from a concentration written with two.

Worth knowing: the number 14.00 belongs to 25 °C alone. Warm the water and it splits a bit more freely. At 37 °C, Kw is about 2.4 × 10-14, so neutral water sits near pH 6.81. Your blood is still neutral by its own temperature, even though the number moved.
Key points
  • Water autoionizes, so every water solution holds both H3O+ and OH-.
  • Kw = [H3O+][OH-] = 1.0 × 10-14 at 25 °C.
  • pH = −log[H3O+] and pH + pOH = 14.00 at 25 °C.
  • One pH unit is a factor of ten in hydronium, not one small step.
Interactive Move pH and watch hydronium and hydroxide trade places. Open full ↑

Section 3Strong versus weak acids and bases

Strong means it goes all the way

A strong acid gives up its proton completely in water. Drop 0.10 mol of HCl into a litre and you get 0.10 mol of H3O+. No HCl molecules survive. The arrow points one way only. The short list is worth memorizing: HCl, HBr, HI, HNO3, HClO4, HClO3 and the first proton of H2SO4.

A strong base works the same way with hydroxide. The soluble metal hydroxides are the main group. LiOH, NaOH and KOH each release one OH-. Ca(OH)2, Sr(OH)2 and Ba(OH)2 each release two.

Worked example · two strong solutions

Find the pH of 0.10 M HCl, then the pH of 0.10 M Ba(OH)2.

Step 1 · the acidFull ionization gives [H3O+] = 0.10 mol/L. So pH = −log(0.10) = 1.00.

Step 2 · the base, per formula unitEach Ba(OH)2 unit sets two hydroxides loose. So [OH-] = 2 × 0.10 = 0.20 mol/L.

Step 3 · pOH then pHpOH = −log(0.20) = 0.70. So pH = 14.00 − 0.70 = 13.30.

CheckThe acid should land low and the base high. 1.00 and 13.30 fit that. Forgetting the factor of two would give pH 13.00, which is wrong by 0.30.

Weak means an equilibrium

A weak acid only partly gives up its proton. Most of it stays whole. Acetic acid in vinegar is the classic case. At 0.10 mol/L it sits at pH 2.87, not 1.00. Only about 1.3 percent of it has ionized. The other 98.7 percent is still CH3COOH.

A weak base behaves the same way. Ammonia takes protons from water, but only a small share of it does. Most of the ammonia stays as NH3. So you write both reactions with ⇌ and solve them as equilibria.

Same concentration, very different amounts of hydronium Each beaker holds 0.10 mol of acid per litre. Eighty acid units are drawn in each. 0.10 M HCl — strong all 80 have split into ions pH 1.00 0.10 M acetic acid — weak only 1 of the 80 has split pH 2.87 H₃O⁺conjugate baseacid molecule still whole
Figure 14.3 Strength is about how far the splitting goes, not how much acid you added. Both beakers hold 0.10 mol of acid per litre. Only the left one turns all of it into hydronium.

Strength and concentration are different ideas

Strength tells you what fraction of the acid splits apart. Concentration tells you how much acid you poured in. They are separate dials. A very dilute strong acid can have a higher pH than a concentrated weak one.

Here is a real example. A 0.0010 M solution of HCl has pH 3.00, while household vinegar at 0.10 M sits at pH 2.87. The weak acid wins on pH here, simply because it brought a hundred times more acid units to the beaker.

The same split shows up with bases. A 0.010 M NaOH solution reaches pH 12.00 because all of its hydroxide is already free. A 0.010 M ammonia solution only reaches about pH 10.6, since most of the ammonia never takes a proton at all.

FeatureStrong acidWeak acid
Arrow→ one way⇌ both ways
Fraction ionizedessentially 100%often under 5%
Intact molecules leftnone you can measuremost of what you added
pH of a 0.10 M sample1.00 for HCl2.87 for acetic acid
Kavery large, not tabulatedsmall, and tabulated
Conjugate basealmost no pull on protonsa real base you must account for
Worth knowing: water sets a ceiling on acid strength. Any acid stronger than H3O+ just converts fully and hands you H3O+ instead. So HCl and HClO4 act the same in water, even though HClO4 is the stronger acid. Chemists call this the leveling effect.
Key points
  • Strong acids and bases ionize fully, so the pH comes straight from the concentration.
  • Weak acids and bases reach an equilibrium with most of the starting material intact.
  • Count hydroxides per formula unit for bases like Ba(OH)2.
  • Strength is a fraction. Concentration is an amount. Never trade one for the other.
Interactive Compare ionization of a strong and a weak acid at the same concentration. Open full ↑

Section 4Ka, Kb and percent ionization

Ka puts a number on weakness

Weak acids need a measure of how far they go. That measure is the acid ionization constant, Ka. It is just the equilibrium constant for the acid giving its proton to water.

Ka = [H3O+][A-] ÷ [HA]

Liquid water is left out, as it is in any equilibrium constant. A bigger Ka means a stronger acid. Formic acid, at 1.8 × 10-4, is ten times stronger than acetic acid at 1.8 × 10-5. Bases get the matching base ionization constant, Kb, built from [BH+][OH-] over [B].

These numbers are easier to compare as logs. pKa is −log Ka. A small pKa means a strong acid. Acetic acid has pKa 4.74, and ammonium has pKa 9.25.

Acid strength ladder, from a strong acid down to very weak ones ACID Ka pKa STRENGTH BAR HCl hydrochloric strong > 1 HSO₄− hydrogen sulfate 1.2 × 10−² 1.92 H₃PO₄ phosphoric 7.5 × 10−³ 2.12 HF hydrofluoric 6.8 × 10−⁴ 3.17 HCOOH formic 1.8 × 10−⁴ 3.74 CH₃COOH acetic 1.8 × 10−⁵ 4.74 H₂CO₃ carbonic 4.3 × 10−⁷ 6.37 NH₄⁺ ammonium 5.6 × 10−¹⁰ 9.25 HCO₃− bicarbonate 4.7 × 10−¹¹ 10.33 Down the list: smaller Ka, larger pKa, weaker acid, stronger conjugate base.
Figure 14.4 A ladder of real acids ranked by Ka. Each rung down is a weaker acid and a stronger conjugate base. The pKa column just flips the exponent into a friendlier number.

Worked example · pH of 0.25 M ammonia

Ammonia has Kb = 1.8 × 10-5. Find the pH of a 0.25 M solution.

Step 1 · the reactionNH3 + H2O ⇌ NH4+ + OH-. Let x be the hydroxide made, in mol/L.

Step 2 · the tableAt balance, [NH4+] = x, [OH-] = x, and [NH3] = 0.25 − x.

Step 3 · solve1.8 × 10-5 = x2 ÷ (0.25 − x). Since Kb is tiny, try 0.25 − x ≈ 0.25. Then x2 = 4.5 × 10-6, so x = 2.1 × 10-3.

Step 4 · pHpOH = −log(2.1 × 10-3) = 2.67. So pH = 14.00 − 2.67 = 11.33.

Check the shortcutx is 0.85 percent of 0.25, well under 5 percent. The approximation was safe. The exact quadratic gives x = 2.11 × 10-3 and pH 11.32.

Ka and Kb are tied together by Kw

An acid and its conjugate base are not independent. Multiply the Ka of an acid by the Kb of its conjugate base. You always get Kw.

Ka × Kb = Kw = 1.0 × 10-14    and    pKa + pKb = 14.00

So one table covers both. Ammonia has Kb = 1.8 × 10-5. Its conjugate acid, ammonium, must have Ka = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10. Acetate runs the same way, with Kb = 5.6 × 10-10.

The rule also explains a pattern. A weaker acid has a stronger conjugate base. Their product is locked, so when one side shrinks the other must grow.

Percent ionization climbs as you dilute

Percent ionization is the ionized fraction times 100. For an acid it is [H3O+] at balance divided by the starting acid concentration. It is not a fixed property. It shifts with concentration.

Acetic acid at 0.10 M is 1.3 percent ionized, giving pH 2.87. Dilute it to 0.0010 M and the quadratic is needed, because x is no longer small next to the start. Solving x2 ÷ (0.0010 − x) = 1.8 × 10-5 gives x = 1.25 × 10-4. That is 12.5 percent ionized, and pH 3.90.

The pH still went up, because there is less acid overall. But the share that split apart grew by almost ten times. Dilution gives the ions more room, so the equilibrium shifts toward them.

Key points
  • Ka and Kb measure strength. Larger means stronger, and smaller pKa means stronger.
  • Ka × Kb = Kw for any conjugate pair, so pKa + pKb = 14.00 at 25 °C.
  • The small-x shortcut is safe when x stays under 5 percent of the starting amount.
  • Percent ionization rises on dilution, even while the pH rises too.
Interactive Change Ka and concentration and track pH and percent ionization. Open full ↑

Section 5Polyprotic acids and salt solutions

Polyprotic acids give protons one at a time

Some acids hold more than one donatable proton. These are polyprotic acids. Carbonic acid has two. Phosphoric acid has three. They never leave all at once. Each proton goes in its own step, with its own Ka.

The steps get much harder as you go. Carbonic acid has Ka1 = 4.3 × 10-7 and Ka2 = 4.7 × 10-11. That is a drop of about 104. The reason is charge. Pulling a positive proton off a particle that is already negative takes much more energy.

Phosphoric acid gives up three protons one step at a time H₃PO₄ H₂PO₄− HPO₄²− PO₄³− − H⁺ Ka1 = 7.5 × 10−³ pKa1 = 2.12 − H⁺ Ka2 = 6.2 × 10−⁸ pKa2 = 7.21 − H⁺ Ka3 = 4.2 × 10−¹³ pKa3 = 12.38 Each step is roughly a hundred thousand times harder than the step before. So the first step sets the pH almost by itself. Pulling H⁺ off a particle that is already negative costs far more energy.
Figure 14.5 Phosphoric acid sheds protons in order, never all at once. The three Ka values drop by about 105 each step. That gap is why you can treat the first step on its own.

This gap is a gift. The second step adds so little hydronium that you can ignore it. So you treat a polyprotic acid as a simple weak acid using Ka1 alone.

Worked example · pH of carbonated water

A soda holds roughly 0.050 M H2CO3. Find its pH using Ka1 = 4.3 × 10-7.

Step 1 · first step onlyH2CO3 + H2O ⇌ HCO3- + H3O+. Let x be the hydronium made.

Step 2 · solve4.3 × 10-7 = x2 ÷ (0.050 − x) ≈ x2 ÷ 0.050. So x2 = 2.15 × 10-8 and x = 1.47 × 10-4.

Step 3 · pHpH = −log(1.47 × 10-4) = 3.83.

Check step twoThe second step would add about 4.7 × 10-11 mol/L of hydronium. Next to 1.47 × 10-4 that is nothing. Dropping it was fair.

Salt solutions are not always neutral

Dissolve a salt and its ions can react with water. That reaction is salt hydrolysis. To predict the result, look at where each ion came from. 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.

Salt built fromExampleReacting ionResult
Strong acid + strong baseNaClneitherneutral, pH 7
Weak acid + strong baseNaCH3COOCH3COO- takes H+basic, pH above 7
Strong acid + weak baseNH4ClNH4+ gives H+acidic, pH below 7
Weak acid + weak baseNH4CH3COOboth ions reactcompare Ka with Kb
Small, high-charge metal ionAlCl3bound water gives H+acidic

Worked example · pH of 0.10 M sodium acetate

Acetic acid has Ka = 1.8 × 10-5. Find the pH of 0.10 M NaCH3COO.

Step 1 · find the active ionSodium came from NaOH, a strong base, so it does nothing. Acetate came from a weak acid, so acetate is a base.

Step 2 · get KbKb = Kw ÷ Ka = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10.

Step 3 · solve for hydroxidex2 ÷ 0.10 = 5.6 × 10-10, so x2 = 5.6 × 10-11 and x = 7.5 × 10-6 mol/L.

Step 4 · pHpOH = −log(7.5 × 10-6) = 5.13. So pH = 14.00 − 5.13 = 8.87.

CheckA salt of a weak acid should come out basic. pH 8.87 is above 7, so the sign is right. Run 0.10 M NH4Cl the same way and you get pH 5.13, the mirror image.

Amphiprotic ions need a comparison

An ion like HCO3- can act either way. So compare its two constants. As an acid it has Ka = 4.7 × 10-11, the Ka2 of carbonic acid. As a base its Kb = Kw ÷ Ka1 = 2.3 × 10-8.

Kb is about 500 times larger, so the base side wins. A baking soda solution turns out basic, which matches the taste and the feel.

Key points
  • Polyprotic acids lose protons in steps, and each Ka is far smaller than the one before.
  • Ka1 sets the pH of a polyprotic acid almost by itself.
  • Ions from strong acids and strong bases are spectators. Ions from weak partners react.
  • For an amphiprotic ion, compare its Ka with its Kb and let the larger one decide.
Interactive Predict whether a salt solution turns out acidic, basic or neutral. Open full ↑

Section 6Buffers and titration curves

A buffer holds pH steady

A buffer mixes a weak acid with its conjugate base, both in real amounts. Add acid and the base half soaks up the extra protons. Add base and the acid half replaces them. Either way the pH barely moves.

Acetic acid with sodium acetate is a standard pair. So is H2PO4- with HPO42-. A strong acid cannot buffer anything, because it keeps no reserve of intact molecules.

The Henderson-Hasselbalch equation

Take the log of the Ka expression and rearrange. You get the Henderson-Hasselbalch equation, the fastest route to a buffer pH.

pH = pKa + log([A-] ÷ [HA])

Two readings fall out at once. Equal amounts of acid and base make the ratio 1, and log 1 is 0, so pH = pKa. And because both terms share one solution, moles work as well as concentrations.

A buffer works best within about one pH unit of its pKa, and that span is called the buffer region. Outside it the ratio gets lopsided and the reserve on one side runs thin. How much strong acid or base a buffer can soak up is its buffer capacity, which grows with the total amount of the pair.

Worked example · a buffer under attack

One litre holds 0.100 mol acetic acid and 0.100 mol sodium acetate. pKa is 4.74. Add 0.010 mol NaOH. Find the pH before and after.

Step 1 · beforeThe ratio is 0.100 ÷ 0.100 = 1. So pH = 4.74 + log(1) = 4.74.

Step 2 · react the strong baseHydroxide eats acid and makes base. Acid falls to 0.090 mol. Acetate rises to 0.110 mol.

Step 3 · afterpH = 4.74 + log(0.110 ÷ 0.090) = 4.74 + 0.09 = 4.83.

CompareThat same 0.010 mol NaOH in one litre of pure water gives [OH-] = 0.010, pOH 2.00 and pH 12.00. The buffer moved 0.09 units. Water moved 5.00.

Titration curves tell the story

A titration curve plots pH against the volume of titrant added. The equivalence point is where the added base exactly matches the acid present.

A strong acid with a strong base gives a flat run, a near vertical jump, then another flat run. Its equivalence point sits at pH 7.00, because the salt left behind is neutral. For 25.00 mL of 0.100 M HCl that jump arrives at 25.00 mL of 0.100 M NaOH.

A weak acid curve differs in three ways. It starts higher, near pH 2.87 for 0.100 M acetic acid. It has a long gentle stretch in the middle, which is the buffer region. And its equivalence point lands above pH 7, because the salt left behind is a base.

Titration curve for a weak acid with a strong base buffer region, pKa ± 1 0 2 4 6 8 10 12 14 0 10 20 30 40 50 volume of 0.100 M NaOH added (mL) pH half-equivalence: pH = pKa = 4.74 equivalence: 25.00 mL, pH 8.72 start pH 2.87
Figure 14.6 Acetic acid titrated with sodium hydroxide, computed point by point. The shaded band is the buffer region. At the half-way mark pH equals pKa, and the equivalence point lands above 7 because acetate is a base.

The half-equivalence point is the most useful spot on the curve. Half the acid has been converted, so acid and base are equal and the ratio is 1. That makes pH equal pKa, which for acetic acid reads 4.74 at 12.50 mL.

At full equivalence every acetic acid molecule has become acetate. That is 2.50 × 10-3 mol in 50.00 mL, so 0.0500 M acetate. Its Kb of 5.6 × 10-10 then gives [OH-] = 5.3 × 10-6 and pH 8.72. A polyprotic acid repeats the pattern, one jump per proton.

Worth knowing: your blood runs on a buffer. Carbonic acid and bicarbonate hold it near pH 7.4. Your lungs adjust the carbonic acid by changing how fast you breathe out CO2. Your kidneys adjust the bicarbonate. Two organs, one Henderson-Hasselbalch ratio.
Key points
  • A buffer needs a weak acid and its conjugate base together in real amounts.
  • pH = pKa + log([A-] ÷ [HA]), and moles work as well as concentrations.
  • Pick a buffer whose pKa is within one unit of the pH you want.
  • At half equivalence pH = pKa. At equivalence a weak acid titration lands above pH 7.
Interactive Titrate acids and bases and watch the curve and the buffer region. Open full ↑

The chapter in one breath

  • An acid donates a proton and a base accepts one, so every transfer makes two conjugate pairs.
  • A free proton in water is really hydronium, H3O+.
  • Water autoionizes, and Kw = 1.0 × 10-14 at 25 °C ties the two ions together forever.
  • pH = −log[H3O+], and pH + pOH = 14.00 at 25 °C.
  • Strong acids and bases ionize fully. Weak ones reach an equilibrium with most material intact.
  • Ka and Kb score that weakness, and Ka × Kb = Kw for any pair.
  • Percent ionization rises as a weak acid gets more dilute.
  • Polyprotic acids lose protons in steps, and Ka1 does nearly all the work on pH.
  • Buffers follow pH = pKa + log([A-]/[HA]), and half equivalence on a curve reads pKa directly.

ReferenceWords worth knowing

Here are the terms this chapter leans on, in alphabetical order.

Acid ionization constant (Ka)
The equilibrium constant for a weak acid giving its proton to water. A larger value means a stronger acid.
Amphiprotic
Able to both donate and accept a proton. Water, HCO3- and H2PO4- all qualify.
Autoionization of water
The slow reaction in which one water molecule hands a proton to another, making H3O+ and OH-.
Base ionization constant (Kb)
The equilibrium constant for a weak base taking a proton from water. A larger value means a stronger base.
Bronsted-Lowry acid and base
An acid is a proton donor. A base is a proton acceptor. The pair of roles defines every reaction in this chapter.
Buffer
A solution of a weak acid and its conjugate base that resists pH change when acid or base is added.
Buffer capacity
How much strong acid or base a buffer can absorb before its pH shifts sharply. It grows with the total amount of the pair.
Conjugate acid-base pair
Two species that differ by one H+. The conjugate base is what an acid becomes after donating, like F- from HF. The conjugate acid is what a base becomes after accepting, like NH4+ from NH3.
Equivalence point
The point in a titration where the added titrant exactly matches the substance being titrated.
Half-equivalence point
Half way to equivalence in a weak acid titration. Acid and conjugate base are equal, so pH equals pKa.
Henderson-Hasselbalch equation
pH = pKa + log([A-]/[HA]). It gives the pH of a buffer from a ratio.
Hydronium ion
H3O+, the real form a donated proton takes in water.
Ion-product constant for water (Kw)
The product [H3O+][OH-]. It equals 1.0 × 10-14 at 25 °C.
Leveling effect
Water limits acid strength. Any acid stronger than H3O+ simply converts fully to H3O+.
Percent ionization
The ionized fraction of a weak acid or base times 100. It rises as the solution gets more dilute.
pH
The negative base-ten logarithm of [H3O+]. Each unit is a factor of ten.
pOH
The negative base-ten logarithm of [OH-]. At 25 °C it is 14.00 minus the pH.
Polyprotic acid
An acid with more than one donatable proton, released in separate steps with separate Ka values.
Salt hydrolysis
The reaction of a dissolved salt ion with water, which can leave the solution acidic or basic.
Titration curve
A plot of pH against the volume of titrant added. Its shape reveals strength and pKa.

Check yourself

1. Find the pH and pOH of 0.0050 M HCl at 25 °C.
pH 2.30 and pOH 11.70. HCl is strong, so [H3O+] = 5.0 × 10-3. pH = −log(5.0 × 10-3) = 2.30. Then pOH = 14.00 − 2.30 = 11.70.
2. Name the conjugate base and the conjugate acid of HSO4-.
SO42- and H2SO4. Remove one H+ to get the conjugate base, SO42-. Add one H+ to get the conjugate acid, H2SO4. HSO4- can do both, so it is amphiprotic.
3. A solution reads pH 3.40. Find [H3O+] and [OH-].
4.0 × 10-4 M and 2.5 × 10-11 M. [H3O+] = 10-3.40 = 4.0 × 10-4. Then [OH-] = (1.0 × 10-14) ÷ (4.0 × 10-4) = 2.5 × 10-11.
4. Find the pH and percent ionization of 0.20 M acetic acid. Ka = 1.8 × 10-5.
pH 2.72 and about 0.95 percent ionized. x2 ÷ 0.20 = 1.8 × 10-5, so x2 = 3.6 × 10-6 and x = 1.9 × 10-3. pH = −log(1.9 × 10-3) = 2.72. Percent = (1.9 × 10-3 ÷ 0.20) × 100 = 0.95.
5. HF has Ka = 6.8 × 10-4. Find Kb for F-, then the pH of 0.10 M NaF.
Kb = 1.5 × 10-11 and pH 8.08. Kb = (1.0 × 10-14) ÷ (6.8 × 10-4) = 1.5 × 10-11. Then x2 = (1.5 × 10-11)(0.10) = 1.5 × 10-12, so [OH-] = 1.2 × 10-6. pOH = 5.92, so pH = 8.08.
6. Find the pH of a buffer that is 0.30 M NH3 and 0.20 M NH4Cl. Kb for NH3 is 1.8 × 10-5.
pH 9.43. Ka for NH4+ = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10, so pKa = 9.25. Then pH = 9.25 + log(0.30 ÷ 0.20) = 9.25 + 0.18 = 9.43.
7. Why is a sodium chloride solution neutral?
Neither ion reacts with water. Na+ comes from NaOH, a strong base, and Cl- comes from HCl, a strong acid. Conjugates of strong species have no pull on protons, so both ions are spectators.
8. Which has the lower pH, 0.10 M HCl or 0.10 M HF? Why?
The HCl, at pH 1.00 against pH 2.10 for the HF. Both have the same concentration, but HCl ionizes fully while HF does not. For HF the quadratic gives [H3O+] = 7.9 × 10-3, so pH = 2.10. Strength, not concentration, separates them here.
9. Is a sodium bicarbonate solution acidic or basic? Back it with numbers.
Basic. As an acid HCO3- has Ka = 4.7 × 10-11. As a base its Kb = (1.0 × 10-14) ÷ (4.3 × 10-7) = 2.3 × 10-8. Kb is far larger, so the base behavior wins.
10. You titrate 25.00 mL of 0.100 M NaOH with 0.100 M HCl. What volume reaches equivalence, and what is the pH there?
25.00 mL, and pH 7.00. Moles of base are (0.100 mol/L)(0.02500 L) = 2.50 × 10-3. Matching that needs the same moles of acid, so 25.00 mL. The product is NaCl, a neutral salt, so pH 7.00.
11. A weak acid is titrated with NaOH. At the half-equivalence point the meter reads 3.74. What is the acid's Ka?
Ka = 1.8 × 10-4. At half equivalence [A-] equals [HA], so pH = pKa = 3.74. Then Ka = 10-3.74 = 1.8 × 10-4. That matches formic acid.
12. Why does pH + pOH = 14.00 hold only at 25 °C?
Because Kw itself changes with temperature. The sum is really −log Kw. Kw is 1.0 × 10-14 only at 25 °C. At 37 °C it is about 2.4 × 10-14, so the sum drops to about 13.62.