In the Brønsted-Lowry model, how is an acid defined?
A proton donor
✓
A proton acceptor
An electron-pair acceptor
A hydroxide-ion donor
Correct Answer
A proton donor
A Brønsted-Lowry acid is a species that gives up a proton (H+) to another species. A proton acceptor is the definition of a Brønsted-Lowry base. An electron-pair acceptor is the Lewis definition of an acid, a different model. A hydroxide-ion donor describes an Arrhenius base, not an acid.
Category: Bronsted-Lowry Acids and Bases
Question 2
What is the conjugate base of H2PO4-?
H3PO4
HPO4 2-
✓
PO4 3-
H2O
Correct Answer
HPO4 2-
A conjugate base is what remains after an acid loses one proton, so removing one H+ from H2PO4- gives HPO4 2-. H3PO4 is the conjugate acid, formed by adding a proton. PO4 3- results from removing two protons. H2O is not related to H2PO4- by the loss of a proton.
Category: Bronsted-Lowry Acids and Bases
Question 3
Which species is the conjugate acid formed when ammonia acts as a base?
NH2-
NH3
NH4+
✓
OH-
Correct Answer
NH4+
A base accepts a proton, and adding one H+ to NH3 gives its conjugate acid, NH4+. NH2- is what ammonia would become by losing a proton, so it is the conjugate base. NH3 is ammonia itself. OH- is formed from water when water donates a proton to ammonia.
Category: Bronsted-Lowry Acids and Bases
Question 4
Which species is amphiprotic, able either to donate or to accept a proton?
Cl-
NH4+
Na+
HCO3-
✓
Correct Answer
HCO3-
HCO3- can donate a proton to become CO3 2- or accept one to become H2CO3, so it is amphiprotic. Cl- has no proton to donate. NH4+ can donate a proton but has no lone pair to accept another. Na+ neither donates nor accepts protons.
Category: Bronsted-Lowry Acids and Bases
Question 5
What is the hydronium ion concentration in pure water at 25 °C?
1.0 × 10^-7 M
✓
1.0 × 10^-14 M
1.0 × 10^-6 M
1.0 × 10^-1 M
Correct Answer
1.0 × 10^-7 M
In pure water, autoionization makes equal amounts of H3O+ and OH-, and Kw = [H3O+][OH-] = 1.0 × 10^-14 at 25 °C, so each concentration is 1.0 × 10^-7 M. 1.0 × 10^-14 M is the value of Kw, not of either ion. 1.0 × 10^-6 M would make water acidic, with a pH of 6. 1.0 × 10^-1 M is the concentration of a fairly concentrated strong acid.
Category: Bronsted-Lowry Acids and Bases
Question 6
A solution has [H3O+] = 1.0 × 10^-2 M at 25 °C. What is its pH?
12.00
2.00
✓
0.010
−2.00
Correct Answer
2.00
pH = −log[H3O+] = −log(1.0 × 10^-2) = 2.00. 12.00 is the pOH of this solution, not its pH. 0.010 is the concentration itself, not its negative logarithm. −2.00 comes from forgetting the negative sign in the definition of pH.
Category: pH and pOH
Question 7
What is the pH of 0.010 M NaOH at 25 °C?
2.00
10.00
12.00
✓
14.00
Correct Answer
12.00
NaOH is a strong base, so [OH-] = 0.010 M, pOH = −log(0.010) = 2.00, and pH = 14.00 − 2.00 = 12.00. 2.00 is the pOH, not the pH. 10.00 does not follow from either logarithm. 14.00 is the sum of pH and pOH at 25 °C, not the pH of this solution.
Category: pH and pOH
Question 8
Which of these acids is a strong acid in water?
HF
CH3COOH
HNO2
HNO3
✓
Correct Answer
HNO3
Nitric acid, HNO3, is one of the common strong acids and ionizes completely in water. HF, hydrofluoric acid, is a weak acid despite being a hydrogen halide. CH3COOH, acetic acid, is a weak acid. HNO2, nitrous acid, is also a weak acid.
Category: Acid and Base Strength
Question 9
Acetic acid has Ka = 1.8 × 10^-5 at 25 °C. What is Kb for the acetate ion at 25 °C?
5.6 × 10^-10
✓
1.8 × 10^-5
5.6 × 10^4
1.8 × 10^-19
Correct Answer
5.6 × 10^-10
For a conjugate acid-base pair, Ka × Kb = Kw, so Kb = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.6 × 10^-10. 1.8 × 10^-5 wrongly assumes the two constants are equal. 5.6 × 10^4 is 1/Ka, which leaves out Kw. 1.8 × 10^-19 comes from multiplying Ka by Kw instead of dividing.
Category: Acid and Base Strength
Question 10
A 0.10 M solution of a weak acid has [H3O+] = 1.3 × 10^-3 M at equilibrium. What is its percent ionization?
0.13%
1.3%
✓
13%
0.013%
Correct Answer
1.3%
Percent ionization = ([H3O+] at equilibrium/initial acid concentration) × 100 = (1.3 × 10^-3/0.10) × 100 = 1.3%. 0.13% and 0.013% come from misplacing the decimal point by one or two places. 13% comes from multiplying by 1000 instead of 100.
Category: Acid and Base Strength
Question 11
A 0.10 M solution of a weak acid HA has Ka = 1.0 × 10^-5. Assuming the ionization is small, what is the pH, to two decimal places?
1.00
5.00
3.00
✓
6.00
Correct Answer
3.00
With [H3O+] = x, x^2/0.10 = 1.0 × 10^-5, so x^2 = 1.0 × 10^-6, x = 1.0 × 10^-3 M and pH = 3.00; x is 1% of 0.10 M, so the assumption holds. 1.00 treats the weak acid as if it ionized completely. 5.00 is the pKa, not the pH. 6.00 comes from forgetting to take the square root of x^2.
Category: Acid and Base Strength
Question 12
Which salt forms a basic solution when dissolved in water?
Sodium chloride
Ammonium chloride
Potassium nitrate
Sodium acetate
✓
Correct Answer
Sodium acetate
The acetate ion is the conjugate base of the weak acid acetic acid, so it accepts protons from water and produces OH-, while Na+ has no effect. Sodium chloride and potassium nitrate are salts of strong acids and strong bases, so their solutions are neutral. Ammonium chloride is acidic because NH4+ donates protons to water.
Category: Hydrolysis of Salts
Question 13
A solution of which metal ion is acidic because the hydrated ion donates protons to water?
Al3+
✓
Na+
K+
Ca2+
Correct Answer
Al3+
The small, highly charged Al3+ ion pulls electron density from the water molecules bound to it, so the hydrated ion releases H+ and the solution becomes acidic. Na+ and K+ are Group 1 ions whose hydrated forms do not measurably ionize. Ca2+ has a lower charge density than Al3+ and has a negligible effect on pH.
Category: Hydrolysis of Salts
Question 14
For a weak diprotic acid H2A, how do the first and second ionization constants, Ka1 and Ka2, compare?
Ka1 is smaller than Ka2
Ka1 is larger than Ka2
✓
Ka1 is equal to Ka2
Ka1 is the square of Ka2
Correct Answer
Ka1 is larger than Ka2
Removing a proton from a neutral molecule is easier than removing one from the negatively charged ion left behind, so each successive ionization constant is smaller and Ka1 exceeds Ka2. Ka1 is smaller than Ka2 reverses the trend. Ka1 is equal to Ka2 ignores the effect of the growing negative charge. Ka1 is the square of Ka2 describes no real relationship.
Category: Polyprotic Acids
Question 15
Which mixture of equal amounts of two compounds in water forms a buffer?
HCl and NaCl
NaOH and NaCl
CH3COOH and CH3COONa
✓
HNO3 and KNO3
Correct Answer
CH3COOH and CH3COONa
A buffer needs a weak acid together with its conjugate base, and acetic acid with sodium acetate supplies both. HCl and NaCl and HNO3 and KNO3 each pair a strong acid with an anion that is too weak a base to remove added acid. NaOH and NaCl contains a strong base and no weak acid to neutralize added base.
Category: Buffers
Question 16
A buffer contains a weak acid with pKa = 4.20 and its conjugate base, and the conjugate base concentration is 10 times the acid concentration. What is the pH?
4.20
3.20
42.0
5.20
✓
Correct Answer
5.20
By the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) = 4.20 + log(10) = 4.20 + 1.00 = 5.20. 4.20 would be the pH only if the two concentrations were equal. 3.20 comes from inverting the ratio. 42.0 comes from multiplying the pKa by the ratio instead of adding its logarithm.
Category: Buffers
Question 17
A small amount of HCl is added to a buffer containing CH3COOH and CH3COO- ions. Which reaction keeps the pH nearly constant?
CH3COO- + H3O+ → CH3COOH + H2O
✓
CH3COOH + OH- → CH3COO- + H2O
Na+ + Cl- → NaCl
H2O + H2O → H3O+ + OH-
Correct Answer
CH3COO- + H3O+ → CH3COOH + H2O
The conjugate base in the buffer reacts with the added hydronium ion, converting it into the weak acid and water, so very little free H3O+ remains. CH3COOH + OH- → CH3COO- + H2O is how the buffer removes added base, not added acid. Na+ + Cl- → NaCl does not occur in solution, because these ions stay dissolved. H2O + H2O → H3O+ + OH- is the autoionization of water and produces hydronium rather than removing it.
Category: Buffers
Question 18
Acetic acid is titrated with NaOH. What is the pH at the equivalence point at 25 °C?
Below 7
Above 7
✓
Exactly 7
Equal to 14
Correct Answer
Above 7
At the equivalence point all the acetic acid has become acetate ion, a weak base that reacts with water to produce OH-, so the pH is above 7. Below 7 is the result for titrating a weak base with a strong acid. Exactly 7 is the result for a strong acid with a strong base. Equal to 14 is far too basic for a solution whose only base is the weak acetate ion.
Category: Titrations
Question 19
A weak acid with Ka = 1.8 × 10^-5 is titrated with NaOH at 25 °C. What is the pH when half of the acid has been neutralized?
7.00
2.37
4.74
✓
9.26
Correct Answer
4.74
At the half-equivalence point the concentrations of the weak acid and its conjugate base are equal, so pH = pKa = −log(1.8 × 10^-5) = 4.74. 7.00 is the pH at the equivalence point of a strong acid-strong base titration. 2.37 is half the pKa. 9.26 is the pKb of the conjugate base, 14.00 − 4.74.
Category: Titrations
Question 20
Which indicator is suitable for titrating a weak acid with a strong base?
Methyl orange
Methyl red
Litmus
Phenolphthalein
✓
Correct Answer
Phenolphthalein
The equivalence point of a weak acid-strong base titration lies above pH 7, and phenolphthalein changes color in roughly the pH 8 to 10 range, which matches the steep rise. Methyl orange and Methyl red change color in acidic solution, well before the equivalence point. Litmus changes over a range that ends near pH 8, below the equivalence point of a typical weak acid titration.
Category: Titrations