3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Acetic acid, CH₃COOH, has a pKa of 4.74. A solution of acetic acid is adjusted to a pH of 6.00 at 25°C by adding NaOH(aq). Which species, CH₃COOH or CH₃COO⁻, is present at the higher concentration in the final solution, and why?
Answer and reasoning
ACH₃COOH, because the pH is higher than the pKa of acetic acid A student who thinks the protonated form predominates when the pH is above the pKa picks this. A pH above the pKa means [H₃O⁺] is below Ka, so the ratio [CH₃COO⁻]/[CH₃COOH] = Ka/[H₃O⁺] is greater than 1.
BCH₃COOH, because the pH of the solution is less than 7 A student who decides the predominant form by whether the solution is acidic or basic picks this. The solution is acidic, but the dividing line for acetic acid is its pKa, 4.74, not 7; at pH 6.00 acetate predominates.
CCH₃COO⁻, because the pH is greater than the pKa of acetic acidCorrect When the pH of a solution is greater than the pKa of an acid, the deprotonated form predominates: here [CH₃COO⁻]/[CH₃COOH] = Ka/[H₃O⁺] = (1.8 × 10⁻⁵)/(1.0 × 10⁻⁶) = 18.
DNeither; the two are equal because the solution is at equilibrium A student who thinks an acid and its conjugate base are present in equal amounts at equilibrium picks this. They are equal only at pH = pKa = 4.74; at pH 6.00, [CH₃COO⁻] is 18 times [CH₃COOH].
Working No calculation. pH 6.00 > pKa 4.74, so [H₃O⁺] = 1.0 × 10⁻⁶ M is less than Ka = 1.8 × 10⁻⁵, and [CH₃COO⁻]/[CH₃COOH] = Ka/[H₃O⁺] = 18 > 1. The acetate ion predominates even though the solution is acidic.
A student wants to predict whether an acid-base indicator, HIn, will be present mainly as HIn or mainly as In⁻ in a particular solution. The pKa of HIn is known. Which additional quantity does the student need?
Answer and reasoning
AThe solution's pHCorrect [In⁻]/[HIn] = Ka/[H₃O⁺], so with the pKa known, the pH of the solution decides which form predominates: HIn below the pKa, In⁻ above it.
BThe molarity of HIn A student who thinks the fraction of an indicator in each form is set by its own concentration, as for an acid alone in water, picks this. In a solution whose pH is set by other solutes, [In⁻]/[HIn] = Ka/[H₃O⁺] does not involve the indicator's concentration.
CThe Kb value of In⁻ A student who thinks the predominant form is decided by comparing Ka of HIn with Kb of In⁻ picks this. Kb = Kw/Ka is already fixed by the known pKa, and the comparison is the same in every solution, so it cannot predict the form in a particular one.
DWhether it's basic A student who thinks every indicator shows its base color in any basic solution picks this. Each indicator changes around its own pKa, so the student needs the pH of the solution, not only whether it is above 7.
Working No calculation. From Ka = [H₃O⁺][In⁻]/[HIn], [In⁻]/[HIn] = Ka/[H₃O⁺]. With Ka known, the only other quantity needed is [H₃O⁺], that is, the pH of the solution: compare pH with pKa.
A student will titrate a solution of a weak acid, HA, with NaOH(aq) at 25°C and must choose an acid-base indicator from several whose pKa values are known. Which value should the student compare with the pKa values of the indicators in order to choose the indicator that gives the most accurate result?
Answer and reasoning
AThe pH of the mixture at the equivalence pointCorrect The indicator's color change is centered near its pKa, so the indicator whose pKa is closest to the pH at the equivalence point changes color as the equivalence point is reached.
BThe pKa of the weak acid that is being titrated A student who thinks the indicator's pKa should match the pKa of the acid being titrated picks this. The pH equals the pKa of HA at the half-equivalence point, so such an indicator would change color when only about half of the acid had reacted.
CThe pH of the mixture once excess NaOH is added A student who thinks the end point is where the titration curve levels off after the steep section picks this. An indicator matched to that pH would change color only after NaOH had been added beyond the equivalence point.
DThe pH of pure water at the same temperature A student who thinks the solution is neutral at the equivalence point of any titration picks this, a pH of 7 at 25°C. At the equivalence point of this titration the solution contains A⁻, which reacts with water to give OH⁻, so the pH there is above 7.
Working An indicator changes color over a pH range centered near its own pKa, and the titration is accurate when that color change coincides with the steep rise in pH at the equivalence point. The quantity needed is therefore the pH at the equivalence point. For HA titrated with NaOH the solution at the equivalence point contains A⁻, which reacts with water to give OH⁻, so that pH is above 7 and is found from Kb of A⁻ and the concentration of A⁻.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.7.A.1 Protonation state Fix
Protonation state
Whether the members of a conjugate acid-base pair are present mainly as the protonated form (HA) or the deprotonated form (A⁻) in a solution; it is described by the ratio [A⁻]/[HA].
Comparing pH with pKa
When the pH of a solution is less than the pKa of an acid, [HA] > [A⁻]; when the pH is greater than the pKa, [A⁻] > [HA]; when pH = pKa, [HA] = [A⁻]. This follows from Ka = [H₃O⁺][A⁻]/[HA], which gives [A⁻]/[HA] = Ka/[H₃O⁺].
Predominant form
The member of a conjugate acid-base pair present at the higher concentration in a solution at a given pH; the other member is still present, at a lower concentration.
Gradual change of [A⁻]/[HA] with pH
Because [A⁻]/[HA] = Ka/[H₃O⁺], each increase of 1 pH unit multiplies [A⁻]/[HA] by 10. One pH unit below the pKa the ratio is 1:10, and one unit above it is 10:1, so the change from mostly HA to mostly A⁻ is gradual rather than a sudden switch at the pKa.
Species-distribution graph
A graph of the percentage of an acid present as HA and as A⁻ against pH. The HA curve falls and the A⁻ curve rises as the pH increases, and the two curves cross at 50% where pH = pKa.
Students often think When the pH of a solution is greater than the pKa of an acid, the protonated form, HA, predominates; when the pH is less than the pKa, A⁻ predominates. In fact The deprotonated form, A⁻. Because [A⁻]/[HA] = Ka/[H₃O⁺], a pH above the pKa means [H₃O⁺] is less than Ka, so [A⁻] > [HA]; a pH below the pKa means [HA] > [A⁻].
Students often think In an acidic solution (pH below 7 at 25°C) an acid is present mainly in its protonated form, and in a basic solution mainly in its deprotonated form. In fact No. It is decided by comparing the solution's pH with the pKa of that acid, not with 7. Acetic acid (pKa 4.74) is present mainly as acetate ion in a solution of pH 6, even though that solution is acidic.
8.7.A.2 Acid-base indicator Fix
Acid-base indicator
A weak acid, written HIn, whose protonated form (HIn) and deprotonated form (In⁻) differ in a property such as color. Because [In⁻]/[HIn] depends on the pH, the color of a solution containing the indicator responds to the pH.
Indicator color change
An indicator changes color over a range of pH values around its pKa, where HIn and In⁻ are both present in significant amounts. Well below the pKa the color of HIn is seen; well above it, the color of In⁻.
Students often think An acid-base indicator shows one color in any acidic solution and its other color in any basic solution, changing at pH 7. In fact No. An indicator shows the color of In⁻ only where the pH is well above its own pKa. An indicator with a pKa of 9.0 is present mainly as HIn in a basic solution of pH 8.
Students often think The fraction of an indicator present as In⁻ is set by the indicator's own concentration and Ka, as for an acid dissolved alone in water, so the concentration of indicator must be known to predict its color. In fact No. A few drops of indicator have a negligible effect on the pH, and the ratio [In⁻]/[HIn] = Ka/[H₃O⁺] depends only on the solution's [H₃O⁺] and the indicator's Ka, not on the amount of indicator.
8.7.A.3 End point Fix
End point
The point in a titration at which the indicator changes color. It is used to locate the equivalence point, at which the amount of titrant added is exactly enough to react completely with the analyte originally present (equal moles for a 1:1 reaction).
Indicator selection
For accurate titration results, an indicator is chosen whose pKa is close to the pH at the equivalence point, so that its color change happens in the steep section of the titration curve at equivalence.
Students often think At the equivalence point of any acid-base titration the solution is neutral, so the indicator should change color at pH 7. In fact No. At 25°C the equivalence point is at pH 7 for a strong acid titrated with a strong base. Titrating a weak acid gives a basic equivalence point, because A⁻ is present, and titrating a weak base gives an acidic one, because HB⁺ is present.
Students often think The best indicator for a titration has a pKa equal to the pKa of the weak acid, or of the conjugate acid of the weak base, being titrated, which is the pH at the half-equivalence point. In fact No. The pKa of the titrated acid (or of the conjugate acid of the titrated base) equals the pH at the half-equivalence point. The indicator's pKa should be close to the pH at the equivalence point, where the curve rises or falls steeply.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The weak acid HA has Ka = 3.0 × 10⁻⁵. A solution containing HA and its conjugate base, A⁻, has a pH of 5.00 at 25°C. Which of the numbered diagrams shown best represents the relative numbers of HA molecules and A⁻ ions in a small volume of the solution? Water molecules and all other ions are not shown.
Answer and reasoning
ADiagram 1 A student who thinks the protonated form predominates when the pH is above the pKa picks this. pKa = −log(3.0 × 10⁻⁵) = 4.52, below the pH of 5.00, and [A⁻]/[HA] = Ka/[H₃O⁺] = 3.0, so A⁻ ions should outnumber HA molecules three to one.
BDiagram 2 A student who thinks an acid and its conjugate base are present in equal amounts at equilibrium picks this. They are equal only when [H₃O⁺] = Ka, at pH 4.52; at pH 5.00, [A⁻]/[HA] = 3.0.
CDiagram 3 A student who thinks an acid is entirely deprotonated once the pH is above its pKa picks this. A⁻ does predominate, but the ratio [A⁻]/[HA] is 3.0, not infinitely large, so a quarter of the acid is still present as HA.
DDiagram 4Correct [H₃O⁺] = 1.0 × 10⁻⁵ M, so [A⁻]/[HA] = Ka/[H₃O⁺] = (3.0 × 10⁻⁵)/(1.0 × 10⁻⁵) = 3.0. The diagram with six A⁻ ions and two HA molecules shows this ratio: A⁻ predominates, and some HA remains.
Working [H₃O⁺] = 10−5.00 = 1.0 × 10⁻⁵ M. From Ka = [H₃O⁺][A⁻]/[HA], [A⁻]/[HA] = Ka/[H₃O⁺] = (3.0 × 10⁻⁵)/(1.0 × 10⁻⁵) = 3.0. Of eight acid particles, 6 should be A⁻ and 2 HA. (pKa = 4.52 is below the pH, so A⁻ predominates, but HA is still present.)
The weak acid HA has a pKa of 5.00. In a solution at 25°C, the pH is 5.60 and the total concentration of the acid in its two forms, [HA] + [A⁻], is 0.100 M. What is [A⁻] in the solution?
Answer and reasoning
A2.0 × 10⁻² M A student who thinks the protonated form predominates when the pH is above the pKa picks this, taking [HA]/[A⁻] = 4.0. The pH is above the pKa, so A⁻ is the larger part: 2.0 × 10⁻² M is [HA], not [A⁻].
B8.0 × 10⁻² MCorrect [H₃O⁺] = 2.5 × 10⁻⁶ M and Ka = 1.0 × 10⁻⁵, so [A⁻]/[HA] = Ka/[H₃O⁺] = 4.0. Four parts of five are A⁻: [A⁻] = 0.100 M × 4.0/5.0 = 8.0 × 10⁻² M.
C5.0 × 10⁻² M A student who thinks an acid and its conjugate base are equal at equilibrium picks this, splitting 0.100 M in half. That is true only at pH = pKa; here [A⁻]/[HA] = 4.0.
D1.0 × 10⁻¹ M A student who thinks an acid is completely deprotonated once the pH is above its pKa picks this. At 0.60 pH unit above the pKa, [A⁻]/[HA] = 4.0, so about 20% of the acid is still HA.
Working [H₃O⁺] = 10−5.60 = 2.5 × 10⁻⁶ M; Ka = 10−5.00 = 1.0 × 10⁻⁵. [A⁻]/[HA] = Ka/[H₃O⁺] = 4.0. With [HA] + [A⁻] = 0.100 M: [A⁻] = 0.100 M × 4.0/5.0 = 8.0 × 10⁻² M, and [HA] = 2.0 × 10⁻² M.
An acid-base indicator, HIn, has a pKa of 9.0 at 25°C; HIn is colorless and In⁻ is pink. A student adds a few drops of the indicator to a solution at 25°C, and the solution stays colorless. The student concludes that the solution is not basic. Which statement best evaluates the student's conclusion?
Answer and reasoning
AIt is not valid, because colorless HIn predominates in solutions with pH above 9.0 A student who thinks the protonated form predominates when the pH is above the pKa picks this. HIn predominates when the pH is below 9.0; above 9.0 the solution is mainly In⁻ and turns pink.
BIt is not valid, because HIn predominates in basic solutions with pH below 9.0Correct Colorless means HIn predominates, so the pH is below the pKa of 9.0. Solutions with pH between 7 and 9.0 are basic at 25°C and still have [HIn] > [In⁻], so the indicator stays colorless in them.
CIt is valid, because HIn predominates only in solutions with pH below 7 A student who thinks an acid is mainly protonated in acidic solutions and mainly deprotonated in basic ones picks this. HIn predominates whenever the pH is below its pKa of 9.0, which includes basic solutions with pH between 7 and 9.0.
DIt is valid, because In⁻ predominates in solutions with pH above 7 A student who thinks every indicator shows its base color in any basic solution picks this. This indicator changes around its pKa of 9.0; in a basic solution of pH 8, for example, [In⁻]/[HIn] = 0.1 and the solution is colorless.
Working No calculation. Colorless means [HIn] > [In⁻], so pH < pKa = 9.0. A solution with pH between 7 and 9.0 is basic at 25°C, yet HIn predominates there, so a colorless result does not show that the solution is not basic.
Glycine contains a carboxylic acid group, −COOH, and an amino group, −NH₂. At 25°C the pKa of its −COOH group is about 2.3, and the pKa of its −NH₃⁺ group (the conjugate acid of −NH₂) is about 9.6. Which of the numbered structures shown is the predominant form of glycine in a solution of pH 6.0?
Answer and reasoning
AStructure 1 A student who thinks acidic groups stay protonated in any acidic solution picks this, because pH 6.0 is below 7. The −COOH group has a pKa of 2.3, well below 6.0, so it is mainly deprotonated, −COO⁻.
BStructure 2 A student who thinks the protonated form predominates when the pH is above the pKa picks this, reversing both comparisons. pH 6.0 is above 2.3, so the carboxyl group is mainly −COO⁻, and below 9.6, so the amino group is mainly −NH₃⁺.
CStructure 3 A student who thinks acidic groups give up their protons completely in water picks this, removing both acidic hydrogens. The −NH₃⁺ group has a pKa of 9.6; at pH 6.0 it is mainly protonated, with [−NH₃⁺]/[−NH₂] ≈ 103.6.
DStructure 4Correct Each group is judged by its own pKa. pH 6.0 is above 2.3, so the carboxylic acid group is mainly deprotonated, −COO⁻; pH 6.0 is below 9.6, so the amino group is mainly protonated, −NH₃⁺.
Working No calculation. Carboxyl group: pH 6.0 > pKa 2.3, so −COO⁻ predominates. Amino group: pH 6.0 < pKa 9.6 of −NH₃⁺, so the protonated −NH₃⁺ predominates. Predominant form: H₃N⁺−CH₂−COO⁻.
A student titrates 25.0 mL of 0.100 M NH₃(aq) with 0.100 M HCl(aq) at 25°C and records the pH with a pH meter. The titration curve plotted from the student's data is shown. The student will repeat the titration using an acid-base indicator in place of the pH meter. For the most accurate result, the pKa of the indicator should be closest to which value?
Answer and reasoning
A5.3Correct The equivalence point is at 25.0 mL, where moles of HCl added equal moles of NH₃. The middle of the steep drop there is at about pH 5.3, acidic because NH₄⁺ is present, so an indicator with a pKa near 5.3 changes color at the equivalence point.
B7.0 A student who thinks every titration has its equivalence point at pH 7 picks this. At the equivalence point of this weak base titration the solution contains NH₄⁺, which makes it acidic, about pH 5.3; pH 7.0 is passed just before 25.0 mL, but the indicator should change color at the equivalence point itself.
C9.3 A student who matches the indicator to the pKa of NH₄⁺, the pH at the half-equivalence point (12.5 mL), picks this. The indicator's pKa should match the pH at the equivalence point, 25.0 mL, which is about 5.3.
D1.6 A student who thinks the end point is where the curve levels off picks this. Where the curve flattens near pH 1.6, HCl is in excess; an indicator changing color there would signal an end point long after the equivalence point.
Working Equivalence point: moles HCl = moles NH₃ = 0.0250 L × 0.100 M = 2.50 × 10⁻³ mol, at 25.0 mL. Reading the curve, the middle of the steep drop at 25.0 mL is at pH ≈ 5.3 (NH₄⁺ makes the solution acidic: [H₃O⁺] = √((1.0 × 10⁻¹⁴/1.8 × 10⁻⁵) × 0.0500) = 5.3 × 10⁻⁶ M, pH 5.28). Choose an indicator with pKa ≈ 5.3.
An acid-base indicator, HIn, is yellow, and its conjugate base, In⁻, is blue. In a solution of pH 5.40 at 25°C, absorbance measurements show that [In⁻] is three times [HIn]. What is the pKa of HIn?
Answer and reasoning
A5.88 A student who puts the un-ionized indicator in the numerator of the Ka expression picks this: Ka = [H₃O⁺][HIn]/[In⁻] = 10−5.40/3. Ka has [In⁻] in the numerator; a pKa above 5.40 would make HIn, not In⁻, the predominant form.
B5.40 A student who thinks the pKa equals the pH of any solution containing the acid picks this. pH = pKa only when [HIn] = [In⁻]; here [In⁻] is three times [HIn].
C4.92Correct Ka = [H₃O⁺][In⁻]/[HIn] = 10−5.40 × 3 = 1.19 × 10⁻⁵, so pKa = 4.92. The pH is above the pKa, consistent with the blue In⁻ form predominating.
D0.12 A student who thinks Ka is the fraction of the indicator in the ionized form picks this: Ka = 3/4, pKa = 0.12. Ka = [H₃O⁺][In⁻]/[HIn] also contains [H₃O⁺].
Working [H₃O⁺] = 10−5.40 = 3.98 × 10⁻⁶ M. Ka = [H₃O⁺][In⁻]/[HIn] = 3.98 × 10⁻⁶ × 3 = 1.19 × 10⁻⁵. pKa = −log(1.19 × 10⁻⁵) = 4.92. Check: pH 5.40 > pKa 4.92, consistent with In⁻ predominating.
The graph shows the percentage of a weak acid present as HA and as A⁻ as a function of pH at 25°C; curves I and II represent the two forms, in some order. Based on the graph, over which pH range is more than 90% of the acid present as HA?
Answer and reasoning
ABelow pH 6.0 A student who thinks an acid is entirely HA at any pH below its pKa picks this. Below pH 6.0, HA predominates, but between pH 5.0 and 6.0 more than 10% of the acid is already A⁻.
BBelow pH 7.0 A student who takes pH 7 as the dividing line for every acid, and so does not read the 90% line on curve I, picks this. Curve I shows only about 10% to 50% HA between pH 6.0 and 7.0; the dividing line is the pKa, 6.0, and HA exceeds 90% only below about pH 5.0.
CAbove pH 7.0 A student who thinks the protonated form predominates at pH values above the pKa picks this, reading curve II as HA. HA is the form that predominates at low pH, so it is curve I, which is above 90% only below about pH 5.0.
DBelow pH 5.0Correct HA predominates at low pH, so curve I, which falls as the pH rises, represents HA. Curve I is above 90% only below about pH 5.0, one unit below the pKa of 6.0, where [A⁻]/[HA] = 1/10.
Working No calculation needed beyond reading the graph. HA predominates at low pH, so curve I, which falls as pH rises, is HA, and curve II is A⁻. The curves cross at 50% at pH 6.0 (= pKa). Curve I is above 90% for pH below about 5.0 ([A⁻]/[HA] = 1/9 at pH = 6.0 − log 9 = 5.05).
A weak base, B, has a pKb of 5.40 at 25°C. At which pH is [HB⁺] equal to four times [B] in an aqueous solution at 25°C?
Answer and reasoning
A8.00Correct The pKa of HB⁺ is 14.00 − 5.40 = 8.60. From Ka = [H₃O⁺][B]/[HB⁺], [H₃O⁺] = Ka × [HB⁺]/[B] = 10−8.60 × 4 = 1.0 × 10⁻⁸ M, so the pH is 8.00, below the pKa of HB⁺ because the acid form predominates.
B9.20 A student who thinks the protonated form predominates when the pH is above the pKa places the pH 0.60 unit above the pKa of HB⁺, at 8.60 + 0.60 = 9.20. At pH 9.20 it is B that is four times as concentrated as HB⁺; the acid form predominates below the pKa, at 8.00.
C4.80 A student who compares the pH directly with the pKb of B, as if it were the pKa of HB⁺, calculates 5.40 − 0.60 = 4.80. The pH must be compared with the pKa of HB⁺, 14.00 − 5.40 = 8.60, which gives 8.60 − 0.60 = 8.00.
D8.60 A student who thinks the pH of any solution containing an acid equals the acid's pKa converts pKb to the pKa of HB⁺, 8.60, and stops there. The pH equals the pKa only when [HB⁺] = [B]; with [HB⁺] four times [B], [H₃O⁺] is four times Ka and the pH is 8.00.
Working The acid form is HB⁺, with pKa = 14.00 − pKb = 14.00 − 5.40 = 8.60, so Ka = 10−8.60 = 2.5 × 10⁻⁹. From Ka = [H₃O⁺][B]/[HB⁺], [H₃O⁺] = Ka × [HB⁺]/[B] = 2.5 × 10⁻⁹ × 4 = 1.0 × 10⁻⁸ M, so pH = 8.00. The pH is below the pKa of HB⁺, as it must be when the acid form has the higher concentration.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
