Chemical and Ionic Equilibrium: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
At 298 K, pKa(C6H5OH)=10.00 and pKb(NH3)=4.75; take pKw=14.00 and 100.25=1.78. The equilibrium constant for C6H5OH(aq)+NH3(aq)⇌C6H5O−(aq)+NH4+(aq) is ______. Exactly one option is correct.
A5.6
B1.8×10−15
C0.18
D1.8×10−5
Show answer and solution
Answer: C
Write the reaction as proton transfer from phenol (acid 1) to ammonia (base 2). Multiply and divide by [H+]:
K=[C6H5OH][NH3][C6H5O−][NH4+]=Ka(NH4+)Ka(C6H5OH).
pKa(NH4+)=14.00−4.75=9.25, so K=10−10.00/10−9.25=10−0.75=1/100.75=100.25/10=0.178≈0.18.
So the equilibrium lies to the left: phenol (pKa 10.00) is a weaker acid than NH4+ (pKa 9.25).
5.6 is the reverse reaction; 1.8×10−15 is KaKb (wrong combination); 1.8×10−5 is Kb alone.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A 1.0×10−4 M aqueous solution of a weak monoprotic acid HA (Ka=1.0×10−10) is prepared at 298 K, where Kw=1.0×10−14. Take log 2 = 0.30. The pH of the solution is ______. Exactly one option is correct.
A7.00
B6.85
C7.15
D6.70
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Answer: B
Here KaC=1.0×10−14 equals Kw, so the H+ from water cannot be ignored.
Charge balance: [H+]=[A−]+[OH−]. Since HA is barely ionised, [HA]≈C and [A−]=KaC/[H+]. Then
[H+]2=KaC+Kw=2.0×10−14, so [H+]=2×10−7 and pH =7−21log2=7−0.15=6.85.
Check: α=Ka/([H+]+Ka)≈7×10−4, so [HA]≈C holds (an exact numerical solution gives pH 6.850).
Traps: KaC alone gives 7.00, which would be a 'neutral' acid solution; adding 10−7 from water to 10−7 from the acid gives 6.70. That double counts, because the acid suppresses water's ionisation.
Difficulty: Peak challenge
Question 3Level 2 · ExamSingle correct
At 298 K, Ksp(BaSO4)=1.0×10−10. Neglect activity effects and any hydrolysis, and take 5=2.236. The molar solubility of BaSO4 in 1.0×10−5 M aqueous Na2SO4 is ______. Exactly one option is correct.
A1.0×10−5 M
B5.0×10−6 M
C3.8×10−6 M
D6.2×10−6 M
Show answer and solution
Answer: D
Let the solubility be s. Then [Ba2+]=s and [SO42−]=s+1.0×10−5.
The usual shortcut (s≪10−5) gives s=10−10/10−5=1.0×10−5 M. That is not small compared with 10−5 and equals the pure-water value, so the shortcut contradicts itself.
Solve exactly: s2+10−5s−10−10=0, so s=2(−1+5)×10−5=0.618×10−5=6.2×10−6 M.
This is less than the pure-water value (1.0×10−5 M), as the common-ion effect requires. 3.8×10−6 (=0.382×10−5) comes from an algebra slip, and 5.0×10−6 from halving.
Difficulty: Hard
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