Please evaluate the final pH of the gastric acid.

Question

Aspirin is a common drug for curing inflammation and fever. The active ingredient of aspirin is known as acetylsalicylic acid (C₉H₈O₄). It is a monoprotic acid and protonates according to the chemical equation below.

C₉H₈O₄ (ap)⇌ H⁺(aq) + C₉H₇O₄^- /(aq)

A nurse has prepared a solution with unknown number of aspirin pills dissolved in 250mL water. Each pill has a mass of 410mg. 50mL of solution requires 24.6mL 0.0925M sodium hydroxide (NaOH) to reach equivalence point.

Please find the number (in integers) of pills used for preparing the 250mL solution.

A patient has ingested the solution prepared by the nurse. As the solution enters into the stomach, 0.8L of gastric acid (HCl) will react with aspirin and caused protonation. Supposed the initial pH of gastric acid is 2.07. Given the equilibrium constant of acid dissociation of aspirin (Ka) is 3x10-5 M.

(b) Please evaluate the final pH of the gastric acid.

Answer 1

(a) 5 pills  |  (b) pH ≈ 2.18

(a) Number of pills

At the equivalence point, moles of NaOH equal moles of aspirin (monoprotic, 1:1):

n(NaOH) = (0.0246 L)(0.0925 M) = 2.276 × 10⁻³ mol — that is what was in the 50 mL aliquot.

Scale to the full 250 mL (a factor of 5): n = 1.138 × 10⁻² mol

Molar mass of C₉H₈O₄ = 180.16 g/mol, so

m = (1.138 × 10⁻²)(180.16) = 2.050 g = 2050 mg

pills = 2050 mg ÷ 410 mg = 5.00 → 5 pills

Landing within 0.1% of a whole number is your confirmation the titration data and the aliquot scaling were handled correctly.

(b) Final pH — this is a common-ion problem

Initial acid in the stomach: [H⁺] = 10^−2.07 = 8.51 × 10⁻³ M
n(H⁺) = (0.800 L)(8.51 × 10⁻³) = 6.81 × 10⁻³ mol

The patient swallows 250 mL, so the total volume becomes 0.800 + 0.250 = 1.05 L:

[H⁺]_{from HCl} = 6.81 × 10⁻³ / 1.05 = 6.48 × 10⁻³ M
[HA]₀ = 1.138 × 10⁻² / 1.05 = 1.084 × 10⁻² M

Now let x be the extra H⁺ released by the aspirin:

K_a = (6.48 × 10⁻³ + x)(x) / (1.084 × 10⁻² − x) = 3 × 10⁻⁵

x ≈ 5.0 × 10⁻⁵ M — under 1% of either term, so the approximation holds.

[H⁺] = 6.48 × 10⁻³ + 5.0 × 10⁻⁵ = 6.53 × 10⁻³ M

pH = −log(6.53 × 10⁻³) = 2.18

The result is more interesting than the number

The pH barely moved — and it went up, not down. Two effects are competing, and neither is large:

• Swallowing 250 mL dilutes the stomach acid, which alone would raise pH to 2.19.
• The aspirin adds H⁺, pulling it back down only to 2.18.

Aspirin contributes almost nothing because it is a weak acid (K_a = 3 × 10⁻⁵) sitting in an already strongly acidic medium. The H⁺ from HCl is the common ion, and it suppresses the aspirin's ionization by Le Châtelier — only about 0.5% of the aspirin ionises. Roughly 11 mmol of aspirin releases just 0.05 mmol of protons.

That suppression is not merely an exam artifact. It is why aspirin is absorbed in the stomach at all: at pH 2 the drug stays in its neutral, un-ionised HA form, which crosses the lipid membrane of the stomach lining far more readily than the charged C₉H₇O₄⁻ anion would.

If your course intends the 250 mL to be neglected and treats the volume as 0.800 L throughout, the same method gives [H⁺] = 8.56 × 10⁻³ M and pH = 2.07 — unchanged to two decimals. Either way the conclusion is the same: aspirin does not meaningfully acidify the stomach. Check which volume convention your problem set expects.