Acids, Bases & Titration — Chapter Practice, Grade 11

hydrogen ion, H⁺ (as H₃O⁺)

Arrhenius acids release H⁺ ions, which combine with water to form hydronium, H₃O⁺.

proton acceptor

A Brønsted-Lowry base accepts a proton (H⁺).

It ionizes completely in water, releasing nearly all its H⁺

A strong acid ionizes essentially completely, releasing almost all of its hydrogen ions into solution.

'Strong' describes how completely the acid ionizes in water — a strong acid ionizes almost 100%. 'Concentrated' describes how much acid is dissolved per volume (a high molarity). The two are independent: you can have a dilute solution of a strong acid or a concentrated solution of a weak acid.

In the first reaction water accepts a proton from HCl (becoming H₃O⁺), so it acts as a Brønsted-Lowry base. In the second reaction water donates a proton to ammonia (becoming OH⁻), so it acts as a Brønsted-Lowry acid. A substance that can do both is called amphoteric.

Acid/base roles depend on the stated reaction.

HCO₃⁻.

State what the supplied observations or conditions establish, and separate that from any assumption.

Yes. It can donate or accept a proton depending on its reaction partner.

No. Their formal concentrations and ionization equilibria differ but give similar hydronium concentration under the model.

Acid strength describes the extent of proton-transfer equilibrium under defined conditions, commonly summarized by Ka. Concentration specifies how much acid is present per solution quantity. A dilute strong acid can have a similar pH to a more concentrated weak acid because they produce similar hydronium activity. The selected comparison uses a monoprotic ideal model at 25°C. Strong does not mean concentrated, corrosive in every context or guaranteed to have the lowest pH among unspecified samples.

No. Their formal concentrations and ionization equilibria differ but give similar hydronium concentration under the model.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

H₂O donates a proton and NH₃ accepts it. Pairs are H₂O/OH⁻ and NH₄⁺/NH₃. NH₃ produces hydroxide through proton transfer with water rather than having to release pre-existing OH⁻ groups.

H₂O donates a proton and NH₃ accepts it. Pairs are H₂O/OH⁻ and NH₄⁺/NH₃. NH₃ produces hydroxide through proton transfer with water rather than having to release pre-existing OH⁻ groups.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the chemical mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

basic

A pH above 7 is basic; 12 is well above 7.

7

At 25°C, Kw≈1.0×10⁻¹⁴ and [H₃O⁺]=[OH⁻]≈1.0×10⁻⁷ M, giving pH≈7. Neutral pH varies with temperature.

5

pH = −log(1 × 10⁻⁵) = 5.

H⁺(aq)+OH⁻(aq)→H₂O(l); spectator ions remain dissolved. Final pH depends on reacting moles and chemical identities. Near 25°C an ideal strong-acid/strong-base mixture is near pH 7 at equivalence, but excess acid or base changes the final pH, and weak-species hydrolysis can change equivalence pH.

1000 times

The activity ratio is 10^(6−3)=1000. This is not automatically a ratio of total acid amounts or titratable acidity.

A concentration between 10⁻² and 10⁻³ M gives pH between 2 and 3.

pH=4.00.

State what the supplied observations or conditions establish, and separate that from any assumption.

pH=11.00, under the stated temperature and dilute approximation.

It is distributed through the final mixed solution under the additive-volume model, not confined to its original portion.

For a selected strong monoprotic acid and strong hydroxide base, first subtract reacting mole amounts using the one-to-one net ratio. Then divide the excess amount by the final solution volume and use the water equilibrium to connect pH and pOH. Averaging the two initial pH values is not a neutralization calculation. Weak acids, polyprotic species and buffering can require additional equilibrium steps. State temperature, extent and volume assumptions before presenting the final number.

It is distributed through the final mixed solution under the additive-volume model, not confined to its original portion.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Acid=0.00200 mol and base=0.00250 mol, leaving 0.00050 mol OH⁻ in 0.0450 L. [OH⁻]≈0.0111 M, pOH≈1.95 and pH≈12.05. Do stoichiometry before logarithms.

Acid=0.00200 mol and base=0.00250 mol, leaving 0.00050 mol OH⁻ in 0.0450 L. [OH⁻]≈0.0111 M, pOH≈1.95 and pH≈12.05. Do stoichiometry before logarithms.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the chemical mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

6

−log 10(10⁻⁶)=6.

No. Strong concerns fraction ionized; concentrated concerns amount per volume.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

4

−log 10(10⁻⁴)=4.

Write the relationship first; substitute with consistent units.

2

−log(.010)=2.

A smaller pK_a corresponds to larger K_a.

The pK_a=3 acid, since its K_a is 100 times larger.

State what the supplied observations or conditions establish, and separate that from any assumption.

x≈√10⁻⁶=0.0010 M; x/C=1%, so small-ionization approximation is reasonable and pH≈3.

Compute x/C, approximately 0.00100/0.100=0.0100, or one percent, and compare with the desired accuracy.

If x is small relative to initial acid concentration C, replace C−x by C to obtain x≈√(KaC). Then check the inferred x/C against a stated tolerance; a classroom five-percent rule is a heuristic, not a universal error guarantee. The selected numbers give about one-percent ionization, so the approximation agrees closely with the quadratic result. When ionization is substantial, added conjugate base is present or water matters, solve the appropriate fuller equilibrium rather than forcing the shortcut.

Compute x/C, approximately 0.00100/0.100=0.0100, or one percent, and compare with the desired accuracy.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the chemical mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

4.5

log 1=0.

The remaining conjugate base accepts protons from water, producing OH⁻.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

8

6+log 100=8.

The conjugate-base component is depleted; the small-addition approximation no longer applies.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

A buffer resists limited additions; it does not have unlimited capacity.

5.0+log10=6.0.

State what the supplied observations or conditions establish, and separate that from any assumption.

The conjugate base produced can react with water to generate OH⁻; equivalence is stoichiometric, not synonymous with neutral pH.

They occupy the same solution volume, so that common denominator cancels in [A⁻]/[HA] under the stated approximation.

The Henderson–Hasselbalch form rearranges the acid equilibrium into a logarithmic conjugate-base/acid ratio. The familiar concentration version approximates activities and often substitutes postreaction component amounts when both species are appreciable. A common solution volume cancels from their mole ratio. Near exhaustion of one component, at very low concentration or under strong nonideality, the simple approximation may fail and a fuller equilibrium is needed. It is not a formula for pure acid before any conjugate base exists.

They occupy the same solution volume, so that common denominator cancels in [A⁻]/[HA] under the stated approximation.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

H⁺ converts A⁻ to HA, leaving 0.100 mol of each. The ratio is one, so pH≈4.80. Assume the usual dilute activity/concentration approximation with both conjugate components appreciable and a common final volume.

H⁺ converts A⁻ to HA, leaving 0.100 mol of each. The ratio is one, so pH≈4.80. Assume the usual dilute activity/concentration approximation with both conjugate components appreciable and a common final volume.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the chemical mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.