Which expression is the solubility product for calcium phosphate, Ca3(PO4)2?
[Ca2+]^3[PO4 3-]^2
✓
[Ca2+]^2[PO4 3-]^3
3[Ca2+] x 2[PO4 3-]
[Ca2+]^3[PO4 3-]^2 / [Ca3(PO4)2]
Correct Answer
[Ca2+]^3[PO4 3-]^2
Ca3(PO4)2 dissolves to give 3 Ca2+ and 2 PO4 3- ions, and each ion concentration is raised to the power of its coefficient. The expression [Ca2+]^2[PO4 3-]^3 swaps the two exponents. The expression 3[Ca2+] x 2[PO4 3-] multiplies by the coefficients instead of using them as exponents. The expression with [Ca3(PO4)2] in the denominator wrongly includes the pure solid, which is left out of every Ksp expression.
Category: Solubility Product
Question 2
Copper(I) bromide, CuBr, has Ksp = 6.3 x 10^-9 at 25 °C. What is its molar solubility in pure water, to two significant figures?
6.3 x 10^-9 M
7.9 x 10^-5 M
✓
1.8 x 10^-3 M
4.0 x 10^-17 M
Correct Answer
7.9 x 10^-5 M
CuBr dissolves into one Cu+ and one Br-, so Ksp = s x s = s^2 and s = the square root of 6.3 x 10^-9, which is 7.9 x 10^-5 M. The value 6.3 x 10^-9 M simply repeats Ksp without taking the root. The value 1.8 x 10^-3 M comes from taking a cube root, which applies to salts that release three ions. The value 4.0 x 10^-17 M squares Ksp instead of taking its square root.
Category: Molar Solubility
Question 3
Find the molar solubility in pure water of calcium fluoride, CaF2, whose Ksp is 4.0 x 10^-11. Round the result to two significant figures.
6.3 x 10^-6 M
3.4 x 10^-4 M
2.2 x 10^-4 M
✓
1.0 x 10^-11 M
Correct Answer
2.2 x 10^-4 M
For CaF2, [Ca2+] = s and [F-] = 2s, so Ksp = s(2s)^2 = 4s^3. Then s^3 = 1.0 x 10^-11 and s = 2.2 x 10^-4 M. The value 6.3 x 10^-6 M treats the salt as if it released only two ions and takes a square root. The value 3.4 x 10^-4 M takes the cube root of Ksp but forgets to divide by 4 first. The value 1.0 x 10^-11 M divides by 4 but never takes the cube root.
Category: Molar Solubility
Question 4
A saturated solution made by dissolving silver sulfate, Ag2SO4, in pure water has [SO4 2-] = 1.4 x 10^-2 M. What is the Ksp of silver sulfate, to two significant figures?
2.7 x 10^-6
3.9 x 10^-4
5.5 x 10^-6
1.1 x 10^-5
✓
Correct Answer
1.1 x 10^-5
Each formula unit releases two Ag+ ions, so [Ag+] = 2.8 x 10^-2 M, and Ksp = [Ag+]^2[SO4 2-] = (2.8 x 10^-2)^2 x (1.4 x 10^-2) = 1.1 x 10^-5. The value 2.7 x 10^-6 uses 1.4 x 10^-2 M for silver ion instead of doubling it. The value 3.9 x 10^-4 multiplies the two concentrations without squaring the silver term. The value 5.5 x 10^-6 squares the sulfate concentration instead of the silver concentration.
Category: Solubility Product
Question 5
In a solution containing the ions of a slightly soluble salt, the ion product Q is found to be greater than the salt's Ksp. What happens next?
Solid forms until Q falls to equal Ksp
✓
More solid dissolves until Q equals Ksp
Nothing changes because the system is at equilibrium
Ksp rises until it equals Q
Correct Answer
Solid forms until Q falls to equal Ksp
When Q exceeds Ksp, the solution holds more dissolved ions than equilibrium allows, so ions combine into solid until Q drops to Ksp. More solid dissolving until Q equals Ksp describes the opposite case, where Q is less than Ksp. Nothing changing because the system is at equilibrium applies only when Q equals Ksp. Ksp rising until it equals Q is impossible because Ksp is a constant at a given temperature.
Category: Predicting Precipitation
Question 6
Two solutions are mixed so that, just after mixing, [Ca2+] = 1.0 x 10^-3 M and [CO3 2-] = 1.0 x 10^-3 M. Calcium carbonate has Ksp = 8.7 x 10^-9. Which statement is correct?
Q = 1.0 x 10^-6, so no precipitate forms
Q = 1.0 x 10^-6, so a precipitate forms
✓
Q = 2.0 x 10^-3, so a precipitate forms
Q = 1.0 x 10^-12, so no precipitate forms
Correct Answer
Q = 1.0 x 10^-6, so a precipitate forms
Q = [Ca2+][CO3 2-] = (1.0 x 10^-3)(1.0 x 10^-3) = 1.0 x 10^-6, which is larger than Ksp = 8.7 x 10^-9, so calcium carbonate precipitates. Q = 1.0 x 10^-6, so no precipitate forms has the right Q but the wrong conclusion, since Q is greater than Ksp. Q = 2.0 x 10^-3 adds the concentrations instead of multiplying them. Q = 1.0 x 10^-12, so no precipitate forms is too small by a factor of one million; 10^-3 times 10^-3 is 10^-6, and that Q is larger than Ksp.
Category: Predicting Precipitation
Question 7
Barium sulfate has Ksp = 1.1 x 10^-10. What is its molar solubility in 0.010 M Na2SO4, assuming the sulfate from the salt itself is negligible?
1.0 x 10^-5 M
1.1 x 10^-12 M
1.1 x 10^-8 M
✓
1.1 x 10^-10 M
Correct Answer
1.1 x 10^-8 M
The sodium sulfate supplies a common ion, so [SO4 2-] is about 0.010 M and Ksp = s(0.010), giving s = 1.1 x 10^-10 / 0.010 = 1.1 x 10^-8 M. The value 1.0 x 10^-5 M is the solubility in pure water, ignoring the common ion. The value 1.1 x 10^-12 M multiplies Ksp by 0.010 instead of dividing. The value 1.1 x 10^-10 M just repeats Ksp.
Category: Common Ion Effect
Question 8
Silver chloride, AgCl, has Ksp = 1.8 x 10^-10, and silver chromate, Ag2CrO4, has Ksp = 1.1 x 10^-12. How do their molar solubilities in pure water compare?
AgCl dissolves to a greater molar extent
The two have equal molar solubilities
They cannot be compared without a third salt
Ag2CrO4 dissolves to a greater molar extent
✓
Correct Answer
Ag2CrO4 dissolves to a greater molar extent
For AgCl, s = the square root of 1.8 x 10^-10 = 1.3 x 10^-5 M; for Ag2CrO4, 4s^3 = 1.1 x 10^-12 gives s = 6.5 x 10^-5 M, so silver chromate is more soluble despite its smaller Ksp. AgCl dissolving to a greater molar extent is the trap of comparing Ksp values for salts with different ion ratios. Equal molar solubilities does not match either calculation. Saying they cannot be compared without a third salt is wrong because each solubility follows directly from its own Ksp.
Category: Molar Solubility
Question 9
Silver nitrate is added slowly to a solution that is 0.010 M in both Cl- and I-. Given Ksp(AgCl) = 1.8 x 10^-10 and Ksp(AgI) = 1.5 x 10^-16, which salt precipitates first and why?
AgI, because it needs a far lower [Ag+] to reach Ksp
✓
AgCl, because it has the larger Ksp value
AgCl, because chloride is the smaller anion
AgI, because iodide is present at higher concentration
Correct Answer
AgI, because it needs a far lower [Ag+] to reach Ksp
AgI begins to form when [Ag+] = 1.5 x 10^-16 / 0.010 = 1.5 x 10^-14 M, while AgCl needs 1.8 x 10^-8 M, so AgI precipitates first; this difference is the basis of selective precipitation. AgCl having the larger Ksp value is a reason it precipitates later, not first. AgCl because chloride is the smaller anion is irrelevant, since ion size does not decide the order. AgI because iodide is present at higher concentration is false because both anions are 0.010 M.
Category: Predicting Precipitation
Question 10
A solution contains 0.0020 M Cl-. Silver chloride has Ksp = 1.8 x 10^-10. What silver ion concentration must be exceeded for AgCl to begin precipitating?
1.3 x 10^-5 M
9.0 x 10^-8 M
✓
3.6 x 10^-13 M
1.8 x 10^-10 M
Correct Answer
9.0 x 10^-8 M
Precipitation starts when [Ag+][Cl-] just exceeds Ksp, so [Ag+] = 1.8 x 10^-10 / 0.0020 = 9.0 x 10^-8 M. The value 1.3 x 10^-5 M is the silver ion concentration in a saturated solution of AgCl in pure water. The value 3.6 x 10^-13 M multiplies Ksp by 0.0020 instead of dividing. The value 1.8 x 10^-10 M treats Ksp itself as the needed concentration.
Category: Predicting Precipitation
Question 11
Boron trifluoride, BF3, readily accepts an electron pair from a donor. What feature of BF3 explains this behavior?
Boron carries a lone pair it can donate
Boron holds an expanded set of ten electrons
Boron has only six valence electrons around it
✓
Fluorine atoms release protons easily
Correct Answer
Boron has only six valence electrons around it
With three B-F bonds and no lone pairs, boron is surrounded by only six electrons, one pair short of an octet, so it accepts an electron pair and acts as a Lewis acid. Boron carrying a lone pair it can donate is false, since boron in BF3 has no lone pair and donating would make it a base. Boron holding an expanded set of ten electrons is false; it has fewer than eight, not more. Fluorine atoms releasing protons is impossible because BF3 contains no hydrogen.
Category: Lewis Acids and Bases
Question 12
Which species is able to act as a Lewis base?
BF3
Al3+
Cu2+
NH3
✓
Correct Answer
NH3
A Lewis base donates an electron pair, and the nitrogen in NH3 has a lone pair available to share. BF3 is electron deficient and acts as a Lewis acid. Al3+ is a metal cation with no lone pairs to give and acts as an electron-pair acceptor. Cu2+ is also a metal cation that accepts electron pairs from ligands, making it a Lewis acid.
Category: Lewis Acids and Bases
Question 13
What distinguishes a coordinate covalent bond from an ordinary covalent bond?
One atom supplies both shared electrons
✓
Each atom supplies one shared electron
Electrons transfer completely to one atom
Three electron pairs are shared
Correct Answer
One atom supplies both shared electrons
In a coordinate covalent (dative) bond, one atom donates both electrons of the shared pair, as when NH3 bonds to H+ to form NH4+. Each atom supplying one shared electron describes an ordinary covalent bond. Electrons transferring completely to one atom describes ionic bonding. Three electron pairs being shared describes a triple bond, which is about bond order rather than electron origin.
Category: Lewis Acids and Bases
Question 14
In the complex ion Cu(CN)2-, formed from Cu+ and CN-, what role does each cyanide ion play?
A central ion that accepts electron pairs
A ligand that donates an electron pair
✓
A spectator ion outside the complex
A proton donor within the complex
Correct Answer
A ligand that donates an electron pair
Each CN- bonds to the copper(I) ion by donating a lone pair, so it is a ligand acting as a Lewis base. A central ion that accepts electron pairs describes Cu+, the Lewis acid of the complex. A spectator ion outside the complex is wrong because the cyanide ions are bonded directly to copper. A proton donor within the complex is wrong because CN- has no hydrogen to give.
Category: Complex Ions
Question 15
The formation constant of Ag(NH3)2+ is Kf = 1.7 x 10^7. What is the dissociation constant Kd of this complex ion, to two significant figures?
1.7 x 10^-7
5.9 x 10^6
5.9 x 10^-8
✓
4.1 x 10^3
Correct Answer
5.9 x 10^-8
Dissociation is the reverse of formation, so Kd = 1/Kf = 1/(1.7 x 10^7) = 5.9 x 10^-8. The value 1.7 x 10^-7 only flips the sign of the exponent without taking the reciprocal of 1.7. The value 5.9 x 10^6 has the correct digits but the wrong power of ten. The value 4.1 x 10^3 is the square root of Kf, which has no meaning here.
Category: Complex Ions
Question 16
Which ion would act as the Lewis acid when it combines with water molecules or ammonia molecules to form a complex ion?
Cl-
F-
OH-
Cu2+
✓
Correct Answer
Cu2+
Transition metal cations such as Cu2+ have empty orbitals that accept electron pairs from ligands, so they are the Lewis acids in complex ions. Cl- is an anion with lone pairs and acts as a Lewis base. F- likewise donates electron pairs, as it does to BF3. OH- donates electron pairs to H+ and to metal ions, making it a Lewis base.
Category: Complex Ions
Question 17
Which substance, added to water in contact with solid calcium fluoride, increases the molar solubility of CaF2?
Nitric acid
✓
Sodium fluoride
Additional solid CaF2
Calcium nitrate
Correct Answer
Nitric acid
Fluoride is the conjugate base of the weak acid HF, so added H+ converts F- to HF, removing fluoride and pulling the dissolution equilibrium toward dissolved ions. Sodium fluoride adds a common ion and lowers the solubility. Additional solid CaF2 does not change the concentrations in a saturated solution. Calcium nitrate adds Ca2+, another common ion that lowers solubility.
Category: Coupled Equilibria
Question 18
For AgBr, Ksp = 5.0 x 10^-13, and for Ag(S2O3)2 3-, Kf = 4.6 x 10^13. What is K for the overall reaction AgBr(s) + 2 S2O3 2-(aq) ⇌ Ag(S2O3)2 3-(aq) + Br-(aq)?
1.1 x 10^-26
23
✓
9.2 x 10^25
4.6 x 10^13
Correct Answer
23
The overall reaction is the sum of AgBr dissolving and Ag+ forming the complex, so its K is the product Ksp x Kf = (5.0 x 10^-13)(4.6 x 10^13) = 23. The value 1.1 x 10^-26 divides Ksp by Kf. The value 9.2 x 10^25 divides Kf by Ksp. The value 4.6 x 10^13 uses only Kf and ignores the dissolution step.
Category: Coupled Equilibria
Question 19
Silver chloride barely dissolves in pure water but dissolves readily in aqueous ammonia. What is the reason?
Ammonia oxidizes chloride to chlorine gas
Ammonia supplies a common ion to the solution
Ammonia binds Ag+ as a complex ion, removing it
✓
Ammonia raises Ksp by warming the solution
Correct Answer
Ammonia binds Ag+ as a complex ion, removing it
Ammonia combines with free Ag+ to form Ag(NH3)2+, which has a large formation constant, so removing Ag+ shifts the AgCl dissolution equilibrium toward dissolved ions. Ammonia oxidizing chloride to chlorine gas is wrong because ammonia is not an oxidizing agent toward chloride. Ammonia supplying a common ion is wrong because NH3 shares no ion with AgCl, and a common ion would lower solubility anyway. Ammonia raising Ksp by warming the solution is wrong because the effect comes from complex formation, not temperature.
Category: Coupled Equilibria
Question 20
Solid aluminum hydroxide, Al(OH)3, dissolves noticeably in which liquid because a soluble complex ion forms?
Pure water
Concentrated NaCl solution
Concentrated Al(NO3)3 solution
Concentrated NaOH solution
✓
Correct Answer
Concentrated NaOH solution
Excess hydroxide bonds to aluminum to form the soluble complex ion Al(OH)4-, so Al(OH)3 dissolves in strong base. Pure water dissolves almost none because Ksp is extremely small. Concentrated NaCl solution supplies no ion that reacts with or complexes aluminum. Concentrated Al(NO3)3 solution adds Al3+, a common ion that lowers solubility.
Category: Coupled Equilibria