Q4. Find the pH of 0.05 M acetic acid (Ka = 1.8 × 10−5) on its own, and compare it with the 4.74 found in the common-ion section when 0.05 M acetate is also present.
[H+] ≈ √(1.8 × 10−5 × 0.05) = √(9.0 × 10−7) = 9.5 × 10−4 M, so pH ≈ 3.02. The common ion cuts [H+] about 53 times, raising the pH from about 3.0 to 4.74.
Q5. Find the pH of 0.10 M sodium acetate (Ka of acetic acid = 1.74 × 10−5).
Acetate hydrolyses: CH3COO− + H2O ⇌ CH3COOH + OH−, with Kb = Kw/Ka = 5.75 × 10−10. [OH−] ≈ √(5.75 × 10−10 × 0.10) = 7.6 × 10−6 M, pOH ≈ 5.12 and pH ≈ 8.88: basic, as expected for a salt of a weak acid and a strong base.
Q6. Arrange in order of increasing acid strength, giving the reason: (a) H2O, H2S, H2Se; (b) PH3, H2S, HCl.
(a) H2O < H2S < H2Se: down group 16 the atom grows and the H–A bond weakens. (b) PH3 < H2S < HCl: across period 3 the electronegativity of A increases, so the H–A bond is more polar.
Q7. The pH of a solution falls from 5 to 3. By what factor has [H+] changed? And [OH−] (298 K)?
[H+] has risen 102 = 100 times (10−5 to 10−3 M). Since [H+][OH−] = 10−14, [OH−] has fallen 100 times (10−9 to 10−11 M).
Q8. At 60 °C, Kw = 9.6 × 10−14. What is the pH of pure water at 60 °C? Is it acidic?
[H+] = √(9.6 × 10−14) = 3.1 × 10−7 M, so pH ≈ 6.51. It is still neutral, because [H+] = [OH−]; “pH 7 is neutral” holds only at 25 °C.