Using the Henderson-Hasselbalch equation with bicarbonate 24 mEq/L, carbonic acid 1.2 mEq/L, pK = 6.1, calculate pH.

Prepare for the BOC Clinical Chemistry Test with our comprehensive quiz. Study with multiple choice questions and detailed explanations. Enhance your readiness and confidence for the exam!

Multiple Choice

Using the Henderson-Hasselbalch equation with bicarbonate 24 mEq/L, carbonic acid 1.2 mEq/L, pK = 6.1, calculate pH.

Explanation:
The Henderson-Hasselbalch equation for the bicarbonate buffer relates pH to the ratio of base to acid: pH = pK + log([HCO3-]/[H2CO3]). With pK 6.1, [HCO3-] = 24 and [H2CO3] = 1.2, the ratio is 24/1.2 = 20. The base- to acid-log is log10(20) ≈ 1.301. Adding to 6.1 gives pH ≈ 7.401, which rounds to 7.40. This shows how changing the bicarbonate to carbonic acid ratio shifts pH; here the ratio yields a pH around 7.40.

The Henderson-Hasselbalch equation for the bicarbonate buffer relates pH to the ratio of base to acid: pH = pK + log([HCO3-]/[H2CO3]). With pK 6.1, [HCO3-] = 24 and [H2CO3] = 1.2, the ratio is 24/1.2 = 20. The base- to acid-log is log10(20) ≈ 1.301. Adding to 6.1 gives pH ≈ 7.401, which rounds to 7.40. This shows how changing the bicarbonate to carbonic acid ratio shifts pH; here the ratio yields a pH around 7.40.

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy