Weak acids, bases, and salts: choose a defensible pH calculation

MCAT content library / Chemistry & physics / Weak acids, bases, and salts: choose a defensible pH calculation

Write the species and reaction first. Then test the shortcut against the change it predicts, and convert the resulting ion concentration to pH.

About 17 minutes, including the worked examples and checks.

From solutions to buffers · Lesson 4 of 6 · Continue in this path

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Concept: choose the equilibrium before choosing the equation
Application: build a concentration account for a weak acid
Changed case: the same acid at a lower concentration
Application: dissolve the conjugate-base salt instead
Added salt changes the initial row
Make the final result answer the chemical question
Check your reasoning
Continue learning
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Concept: choose the equilibrium before choosing the equation

A weak-acid pH calculation begins with the species actually dissolved. A solution of HA, a solution of its salt NaA, and a mixture containing both can share the same acid–base family while requiring different starting conditions. AAMC category 5A includes weak dissociation, added-salt effects, salt hydrolysis, and pH calculations.¹

Use ideal aqueous solutions at 25°C throughout this lesson, with K_w = 1.0 × 10⁻¹⁴. Each example has one relevant proton-transfer step. The numerical systems are original illustrative cases. The aim is to select a justified model and check its result, rather than to apply the square-root shortcut to anything labeled “weak.”

For a weak acid, HA + H₂O ⇌ H₃O⁺ + A⁻, use K_a. For a weak base, B + H₂O ⇌ BH⁺ + OH⁻, use K_b. The unknown generated ion is hydronium in the first model and hydroxide in the second. That distinction determines whether the first logarithm gives pH or pOH.²

Application: build a concentration account for a weak acid

Suppose HA has K_a = 2.5 × 10⁻⁵ and is dissolved at an analytical concentration of 0.0400 M, with no added A⁻. Let x be the concentration that ionizes. Provisionally neglect hydronium from water; verify afterward that x is large compared with 10⁻⁷ M.

Stage	HA	H₃O⁺	A⁻
Initial model	0.0400 M	Neglected	0
Change	−x	+x	+x
Equilibrium	0.0400 − x	x	x

Initial, change, and equilibrium concentrations for HA

The equilibrium expression is K_a = x²/(0.0400 − x). If x is small relative to 0.0400, replace only the denominator’s 0.0400 − x with 0.0400. The numerator remains x²: treating x as small does not mean setting the newly generated ions to zero.³

This gives x ≈ √(K_aC) = √[(2.5 × 10⁻⁵)(0.0400)] = 1.0 × 10⁻³ M. Therefore pH ≈ 3.00. The estimated change is x/C = 0.0010/0.0400 = 0.025, or 2.5%. It is small, and the hydronium estimate is much greater than water’s 10⁻⁷ M scale.

For the teaching estimates here, use 5% as a practical checkpoint for neglecting a depletion term. It is a tolerance choice, not the definition of a weak acid or a universal accuracy guarantee. A tighter numerical task can require a tighter tolerance. Solving the unapproximated equation in this example gives x ≈ 9.88 × 10⁻⁴ M and pH ≈ 3.005, confirming that the shortcut was adequate for the stated estimate.

Changed case: the same acid at a lower concentration

Now decrease C to 2.50 × 10⁻⁴ M without changing K_a. The shortcut predicts x ≈ 7.91 × 10⁻⁵ M, which is 31.6% of C. Calling the acid weak cannot justify ignoring that depletion. The approximation fails its own check even though K_a remains much smaller than one.

Retain C − x and rearrange K_a = x²/(C − x) to x² + K_ax − K_aC = 0. The physically meaningful root is x = [−K_a + √(K_a² + 4K_aC)] / 2. It gives about 6.75 × 10⁻⁵ M and pH 4.17. The other root is negative and cannot represent a concentration.³

Dilution reduced the hydronium concentration but increased the fraction ionized. These changes are compatible: a larger fraction of a much smaller analytical amount can still produce fewer hydronium ions per liter. This is why neither “weak” nor “dilute” alone selects a numerical approximation.

Application: dissolve the conjugate-base salt instead

Prepare a separate 0.0400 M solution of NaA, where A⁻ is the conjugate base of the same HA. Assume complete salt dissociation and negligible acid–base chemistry from Na⁺. Dissolution gives A⁻; it does not initially give an equal concentration of HA. A⁻ then accepts a proton from water:

A⁻ + H₂O ⇌ HA + OH⁻.

For this conjugate pair, K_b = K_w/K_a = 10⁻¹⁴/(2.5 × 10⁻⁵) = 4.0 × 10⁻¹⁰. This hydrolysis is what makes the salt solution basic. A salt made from a weak base’s conjugate acid can instead be acidic; the word “salt” does not imply neutral pH.⁴

Stage	A⁻	HA	OH⁻
Initial model	0.0400 M	0	Neglected
Change	−y	+y	+y
Equilibrium	0.0400 − y	y	y

Conjugate-base hydrolysis: define a new unknown y

Using the small-change model, y ≈ √[(4.0 × 10⁻¹⁰)(0.0400)] = 4.0 × 10⁻⁶ M. Its depletion fraction is only 0.010%, so that approximation is excellent. Here y is hydroxide: pOH ≈ 5.40, and pH = 14.00 − 5.40 = 8.60. Converting hydroxide to pH requires this extra step at 25°C.⁵

Added salt changes the initial row

Finally, suppose a preparation contains 0.0400 M HA and 0.0200 M added NaA after mixing. The acid expression becomes K_a = x(0.0200 + x)/(0.0400 − x), not x²/(0.0400 − x). The conjugate base already exists before the additional acid ionization.

If x is small compared with both supplied concentrations, x ≈ K_a(0.0400)/(0.0200) = 5.0 × 10⁻⁵ M, giving pH ≈ 4.30. The estimated changes are 0.125% of HA and 0.250% of A⁻, supporting the approximation. Added conjugate base has suppressed further HA ionization. Deliberately combining these partners leads to buffer calculations, where the next task is tracking how added strong acid or base changes their amounts.

Make the final result answer the chemical question

Before reporting pH, identify which ion the unknown represents, test every neglected change, and check that the result agrees with the dissolved species’ acid–base behavior. If the predicted hydronium or hydroxide is comparable to water’s own ionization scale, include K_w and the charge and material balances rather than treating water’s contribution as zero. Likewise, a salt with two reactive ions requires consideration of both; the single-equilibrium examples here do not establish its pH by themselves.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A neutral weak base B has K_b = 2.0 × 10⁻⁵ in water at 25°C. Its salt BHCl dissolves completely into BH⁺ and Cl⁻; chloride has negligible acid–base effect. Which equilibrium and constant describe the appreciable acid–base reaction of the dissolved cation?

A. B + H₂O ⇌ BH⁺ + OH⁻; K = 2.0 × 10⁻⁵

Review this choice. This describes the neutral base B, not the initially supplied BH⁺. The cation is B’s conjugate acid and donates a proton to water.

B. BH⁺ + H₂O ⇌ B + H₃O⁺; K = 5.0 × 10⁻¹⁰

Correct. BH⁺ acts as an acid. Its K_a is K_w/K_b = 1.0 × 10⁻¹⁴/(2.0 × 10⁻⁵) = 5.0 × 10⁻¹⁰.

C. Cl⁻ + H₂O ⇌ HCl + OH⁻; K = 2.0 × 10⁻⁵

Review this choice. The prompt states that chloride has negligible acid–base effect. The supplied K_b belongs to B and cannot be transferred to an unrelated ion.

D. BH⁺ + H₂O ⇌ B + H₃O⁺; K = 2.0 × 10⁻⁵

Review this choice. The equilibrium is correctly chosen, but its constant is Kₐ of BH⁺. K_b of B must be converted using the conjugate relationship.

The principle to carry forward

Dissociate the salt first, identify the reactive ion, and convert the constant for that ion’s conjugate pair.

Check 2

An original ideal-solution example contains 0.0250 M monoprotic HA with K_a = 1.0 × 10⁻⁵ at 25°C and no added conjugate base. Using a checked small-change approximation, which pH is closest?

A. 1.60

Review this choice. This treats the entire 0.0250 M acid as hydronium, as though ionization were complete. The equilibrium instead gives about 5.0 × 10⁻⁴ M hydronium.

B. 3.30

Correct. x ≈ √[(1.0 × 10⁻⁵)(0.0250)] = 5.0 × 10⁻⁴ M. This is 2.0% of the starting acid and far above water’s contribution. −log(5.0 × 10⁻⁴) ≈ 3.30.

C. 5.00

Review this choice. This is pKₐ, not the pH of a solution initially containing only the weak acid. The acid concentration also enters the equilibrium calculation.

D. 10.70

Review this choice. This subtracts 3.30 from 14 as though the generated ion were hydroxide. The written acid equilibrium generates hydronium directly.

The principle to carry forward

The square-root estimate is useful only after identifying the reaction and checking both solute depletion and water’s contribution.

Check 3

For another monoprotic acid, C = 0.00200 M and K_a = 5.0 × 10⁻⁴. A student uses x ≈ √(K_aC) and obtains x = 0.00100 M. Water’s contribution is negligible. What should the student do next?

A. Accept the estimate because a constant smaller than one always makes reactant depletion negligible.

Review this choice. A constant must be interpreted relative to the initial concentration and resulting change. Here the estimated depletion is half of the starting acid.

B. Accept the estimate because taking a logarithm removes the need to check the concentration approximation.

Review this choice. A logarithm can compress numerical differences but does not justify an incorrect equilibrium setup. The neglected depletion is substantial.

C. Retain C − x and solve the equilibrium equation because the estimated depletion is 50% of the starting acid.

Correct. x/C = 0.00100/0.00200 = 0.50. Retaining the depletion term gives x² + (5.0 × 10⁻⁴)x − 1.0 × 10⁻⁶ = 0, with a positive root about 7.81 × 10⁻⁴ M.

D. Set x equal to C because failure of the small-change approximation establishes complete ionization.

Review this choice. The failure only rejects the shortcut. It does not replace the weak-equilibrium model with complete ionization; the positive quadratic root remains below C.

The principle to carry forward

Check the approximation against its own calculated change. If it fails, retain the term you neglected rather than changing the chemical identity.