Question 1
At 25 °C, a solution has a pH of 5. What is its pOH?
- A.
2
- B.
5
- C.
9
- D.
10
Answer: C. 9
At 25 °C, pH + pOH = 14, so pOH = 14 − 5 = 9.
MCAT · Chemical and Physical Foundations · General Chemistry
Equilibrium, acids and bases
Equilibrium, acids and bases is part of General Chemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include henderson-Hasselbalch and buffer behavior, weak acid-strong base titration curve, le Chatelier's principle and reaction quotient versus K.
Common traps: inverting the ratio; treating the buffer as unchanged after adding a strong acid; using it outside the useful pKa plus or minus one range.
Sample questions
Question 2
A food chemist prepares a 0.10 M solution of a new weak acid preservative and measures its pH as 3.0. What is the Ka of the acid?
- A.
1.0 × 10⁻⁶
- B.
1.0 × 10⁻⁵
- C.
1.0 × 10⁻³
- D.
1.0 × 10⁻²
Answer: B. 1.0 × 10⁻⁵
With [H⁺] = [A⁻] = 1.0 × 10⁻³ M and [HA] ≈ 0.10 M, Ka = (1.0 × 10⁻³)²/0.10 = 1.0 × 10⁻⁵ (B). A squares [H⁺] but forgets to divide by the acid concentration, C reports [H⁺] itself as Ka, and D divides [H⁺] by the concentration without squaring.
- A.
Question 3
Which oxoacid of chlorine is the weakest acid in water?
- A.
HClO
- B.
HClO₂
- C.
HClO₃
- D.
HClO₄
Answer: A. HClO
Each additional oxygen on chlorine withdraws electron density and spreads the negative charge of the conjugate base over more atoms, so acid strength rises from HClO to HClO₄; HClO is the weakest (A). B and C each carry more oxygens than HClO, and D is a strong acid, the opposite extreme.
- A.
Question 4
A chemist dissolves the sodium salt of a carboxylic acid (Ka = 1.0 × 10⁻⁵) in water to make a 0.10 M solution at 25 °C. What is the pH of the solution?
- A.
3.0
- B.
5.0
- C.
7.0
- D.
9.0
Answer: D. 9.0
The carboxylate is a weak base with Kb = Kw/Ka = 1.0 × 10⁻⁹, so [OH⁻] = √(10⁻⁹ × 0.10) = 10⁻⁵ M, pOH = 5.0 and pH = 9.0 (D). A treats the solution as the acid itself, B stops at pOH, and C assumes every salt is neutral.
- A.
Question 5
A 1.0 L phosphate buffer contains 0.15 mol H₂PO₄⁻ and 0.15 mol HPO₄²⁻ (pKa = 7.2). After 0.050 mol of HCl is added, what is the pH? (log 2 = 0.30)
- A.
1.3
- B.
6.9
- C.
7.2
- D.
7.5
Answer: B. 6.9
HCl converts 0.050 mol HPO₄²⁻ into H₂PO₄⁻, leaving 0.10 mol base and 0.20 mol acid, so pH = 7.2 + log(0.10/0.20) = 7.2 − 0.30 = 6.9 (B). A ignores the buffer and treats the HCl as if in pure water, C assumes the buffer is unchanged, and D inverts the ratio.
- A.
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