4.8 Introduction to Acid-Base Reactions

1. AHSO₄⁻, NH₃
   A student who thinks the acid's conjugate partner is the base it reacts with picks this. HSO₄⁻ is the acid, but its conjugate base is the species it becomes after losing H⁺, SO₄²⁻, not NH₃.
2. BHSO₄⁻, NH₄⁺
   A student who thinks a conjugate pair is the two acids in the equation, one on each side, picks this. HSO₄⁻ is the acid, but NH₄⁺ differs from it by far more than one proton; the conjugate base of HSO₄⁻ is what remains after it loses H⁺, SO₄²⁻.
3. CHSO₄⁻, SO₄²⁻ Correct
   HSO₄⁻ gives a proton to NH₃, so HSO₄⁻ is the acid; what is left, SO₄²⁻, has one H fewer and one more negative charge, so it is the conjugate base of HSO₄⁻.
4. DNH₃, NH₄⁺
   A student who thinks the acid is the species that gains the proton picks this. NH₃ accepts the proton, so it is the base, and NH₄⁺ is its conjugate acid; the proton donor, HSO₄⁻, is the acid.

CED 4.8.A.1 · Read this in Fix

1. AView 1
   A student who thinks any species containing H acts as an acid in water picks this, with NH₃ giving a proton to water to form NH₂⁻ and H₃O⁺. In water NH₃ accepts a proton instead, from a water molecule, forming NH₄⁺ and OH⁻.
2. BView 2 Correct
   In water a small fraction of the NH₃ molecules accept protons from water molecules: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The solution therefore holds mostly NH₃ molecules with equal numbers of NH₄⁺ and OH⁻ ions, as in this view.
3. CView 3
   A student who thinks ammonia and water combine into NH₄OH molecules picks this. No NH₄OH molecules form; NH₃ takes a proton from water, giving separate NH₄⁺ and OH⁻ ions, and most NH₃ stays as molecules.
4. DView 4
   A student who thinks a base must contain OH, so NH₃ cannot react as a base, picks this view with no ions. NH₃ is a Brønsted-Lowry base: it accepts protons from water, so some NH₄⁺ and OH⁻ ions are present.

CED 4.8.A.2 · Read this in Fix

1. AHF and F⁻; H₃O⁺ and H₂O Correct
   HF loses one H⁺ and becomes F⁻, so HF and F⁻ differ by one proton; H₂O gains that H⁺ and becomes H₃O⁺, so H₃O⁺ and H₂O differ by one proton. Each conjugate pair is a reactant with the product that differs from it by one H⁺.
2. BHF and H₂O; F⁻ and H₃O⁺
   A student who thinks a conjugate pair is the acid and base that react with each other picks this. HF and H₂O differ by more than a proton; HF's partner is F⁻, the species it becomes after giving up H⁺.
3. CHF and H₃O⁺; H₂O and F⁻
   A student who thinks a conjugate pair is the two acids in the equation (and the other pair the two bases) picks this. HF and H₃O⁺ do not differ by one proton; each acid is paired with the base that differs from it by one H⁺.
4. DHF and F⁻; H₂O unpaired
   A student who thinks water is only the solvent picks this. The diagram shows the water molecule gaining the proton and becoming H₃O⁺, so H₃O⁺ and H₂O form the second conjugate pair.

CED 4.8.A.3 · Read this in Fix

1. AX⁻ is the stronger base, because HX molecules transfer protons to water to a larger extent than HY molecules do.
   A student who thinks a stronger acid has a stronger conjugate base picks this. The observation is right, but it points the other way: X⁻ takes back protons so weakly that HX ionizes completely, so X⁻ is the weaker base.
2. BY⁻ and X⁻ are equally strong bases, because the two acid solutions were prepared with the same concentration.
   A student who equates strength with concentration picks this. Equal concentrations mean equal amounts of acid dissolved, but the bulbs and particulate views show that HX ionizes far more than HY, so the acids, and their conjugate bases, differ in strength.
3. CX⁻ is the stronger base, because the HY solution contains more un-ionized acid molecules than the HX solution does.
   A student who thinks more un-ionized molecules means more acid, so HY is the stronger acid, picks this. More HY molecules remaining means HY has transferred fewer protons, so HY is the weaker acid and Y⁻ the stronger base.
4. DY⁻ is the stronger base, because HY molecules transfer protons to water to a smaller extent than HX does. Correct
   The bright bulb and the particulate view both show that HX has given all its protons to water while most HY molecules keep theirs. An acid that gives up its proton less readily has a conjugate base that holds on to a proton more strongly, so Y⁻ is the stronger base.

CED 4.8.A.3 · Read this in Fix

1. AHF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l) Correct
   HF is a weak acid, so it is written as molecules; NaOH supplies OH⁻, which takes the proton from HF. Na⁺ is a spectator ion. The equation is balanced in atoms and in charge (−1 on each side).
2. BH₃O⁺(aq) + OH⁻(aq) → 2 H₂O(l)
   A student who thinks every neutralization has the net ionic equation H₃O⁺ + OH⁻ → 2 H₂O picks this. HF is a weak acid, present mainly as HF molecules, so the proton is taken from HF itself and HF appears in the net ionic equation.
3. CHF(aq) + NaOH(aq) → NaF(aq) + H₂O(l)
   A student who thinks every acid-base reaction is written as acid + base → salt + water picks this molecular equation. NaOH and NaF are soluble ionic compounds present as ions; Na⁺ appears on both sides and is removed from the net ionic equation.
4. DHF(aq) + Na⁺(aq) → NaF(aq) + H⁺(aq)
   A student who thinks the metal ion of the base replaces the hydrogen of the acid picks this. The proton goes to OH⁻, forming water; Na⁺ is a spectator ion and does not react.

Working NaOH is a soluble ionic compound: Na⁺(aq) + OH⁻(aq). HF is a weak acid, present mainly as HF molecules, so it is written as HF(aq). The proton moves from HF to OH⁻: HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l). Na⁺ is a spectator ion and is omitted. Charge: −1 = −1; atoms: H 2, F 1, O 1 on each side.

CED 4.8.A.1 · Read this in Fix

1. AHSO₄⁻(aq) + H₂O(l) → SO₄²⁻(aq) + H₃O⁺(aq)
   A student who thinks the acid is the species that gains the proton picks this. Water gains the proton and becomes H₃O⁺, so here water is the base and HSO₄⁻ is the acid.
2. BNH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l)
   A student who thinks the product formed by gaining a proton is the acid picks this, because H₂O forms when OH⁻ gains H⁺. That makes H₂O the conjugate acid, a product; in the direction written, H₂O is a product and donates nothing. The acid of the reaction is NH₄⁺.
3. CNH₃(aq) + H₃O⁺(aq) → NH₄⁺(aq) + H₂O(l)
   A student who treats H₃O⁺ as simply water picks this. H₃O⁺ is the proton donor here, and H₂O is formed as its conjugate base; H₂O itself does not give up a proton.
4. DCO₃²⁻(aq) + H₂O(l) → HCO₃⁻(aq) + OH⁻(aq) Correct
   Here a water molecule is a reactant that gives up H⁺: it becomes OH⁻ while CO₃²⁻ gains the proton and becomes HCO₃⁻. Water is the proton donor, the Brønsted-Lowry acid.

Working Water acts as an acid when an H₂O molecule is a reactant that loses H⁺. In CO₃²⁻ + H₂O → HCO₃⁻ + OH⁻, H₂O gives a proton to CO₃²⁻ and becomes OH⁻. In HSO₄⁻ + H₂O → SO₄²⁻ + H₃O⁺, H₂O gains a proton (base). In NH₄⁺ + OH⁻ → NH₃ + H₂O, H₂O is a product (conjugate acid of OH⁻). In NH₃ + H₃O⁺ → NH₄⁺ + H₂O, H₃O⁺ is the acid and H₂O a product.

CED 4.8.A.2 · Read this in Fix

1. ANH₄OH
   A student who thinks NH₄⁺ and OH⁻ combine into NH₄OH molecules picks this. When OH⁻ takes the proton from NH₄⁺, the products are separate NH₃ and H₂O molecules.
2. BNaCl + H₂O
   A student who thinks every acid-base reaction produces a salt and water picks this. Na⁺ and Cl⁻ stay in solution as spectator ions, so NaCl does not appear in the net ionic equation, and NH₃ is formed.
3. CNH₃ + H₂O Correct
   NH₄⁺ gives a proton to OH⁻: NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l). Na⁺ and Cl⁻ are spectator ions, so the products of the net ionic equation are NH₃ and H₂O.
4. DH₂O
   A student who thinks the net ionic equation of every neutralization is H₃O⁺ + OH⁻ → 2 H₂O picks this. The proton donor here is NH₄⁺, not H₃O⁺, so NH₃ is formed along with water.

Working Ions present: NH₄⁺, Cl⁻, Na⁺, OH⁻. NH₄⁺ donates a proton to OH⁻: NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l). Na⁺ and Cl⁻ are spectator ions. Products: NH₃ and H₂O.

CED 4.8.A.1 · Read this in Fix

1. AA larger volume is needed for the HX solution.
   A student who thinks OH⁻ reacts only with the H₃O⁺ already present picks this. As OH⁻ removes protons, HY molecules keep giving up protons, so all 0.0025 mol of HY react and the volumes are equal.
2. BThe volumes needed for the two acids are equal. Correct
   Each flask holds (0.0250 L)(0.10 M) = 0.0025 mol of monoprotic acid. OH⁻ takes the proton from H₃O⁺ in the HX solution and directly from HY molecules in the HY solution, so both acids give up one proton per molecule and need the same amount, and volume, of NaOH.
3. CThe HY solution needs a larger volume.
   A student who thinks the solution with more un-ionized acid molecules contains more acid picks this. Both flasks contain 0.0025 mol of acid; HY simply has more of it as molecules, which still give one proton each to OH⁻.
4. DNo volume of NaOH brings HY to equivalence.
   A student who thinks an acid that ionizes only slightly does not react with bases picks this. HY transfers few protons to water, but OH⁻ is a much stronger base and takes its protons until all the HY has reacted.

CED 4.8.A.1 · Read this in Fix

1. AH₂PO₄⁻
   A student who thinks a negative ion cannot donate a proton picks this. H₂PO₄⁻ has transferable H atoms and can give a proton to a base, forming HPO₄²⁻.
2. BNH₄⁺
   A student who thinks an ion formed from a base keeps that base's properties picks this. NH₄⁺ is the conjugate acid of NH₃ and can give a proton to a base such as OH⁻.
3. CCO₃²⁻ Correct
   A Brønsted-Lowry acid must donate a proton, and CO₃²⁻ has no hydrogen atom to give. It can act only as a base, accepting protons to form HCO₃⁻.
4. DH₂O
   A student who thinks water is neutral and so cannot act as an acid picks this. Water donates a proton to bases such as NH₃ and CO₃²⁻, forming OH⁻.

CED 4.8.A.1 · Read this in Fix