An increase in [HCO3-] of 10 mEq/L will result in an increase in pH by approximately how much?

An increase in [HCO3-] of 10 mEq/L will result in an increase in pH by approximately how much?

• A. 0.10 pH units
• B. 0.15 pH units
• C. 0.20 pH units
• D. 0.25 pH units

Answer: (B)

The pH change from a shift in bicarbonate hinges on Henderson-Hasselbalch: pH = 6.1 + log([HCO3-] / (0.03 × PaCO2)). If PaCO2 is roughly 40 mmHg (so the denominator is about 1.2), increasing [HCO3-] by 10 mEq/L changes the ratio from 24/1.2 to 34/1.2. That is a log change from log(20) ≈ 1.301 to log(28.333) ≈ 1.452, giving a pH increase of about 0.151. So about 0.15 pH units. This assumes PaCO2 remains constant; changes in CO2 would alter the amount of pH change.

The pH change from a shift in bicarbonate hinges on Henderson-Hasselbalch: pH = 6.1 + log([HCO3-] / (0.03 × PaCO2)). If PaCO2 is roughly 40 mmHg (so the denominator is about 1.2), increasing [HCO3-] by 10 mEq/L changes the ratio from 24/1.2 to 34/1.2. That is a log change from log(20) ≈ 1.301 to log(28.333) ≈ 1.452, giving a pH increase of about 0.151. So about 0.15 pH units. This assumes PaCO2 remains constant; changes in CO2 would alter the amount of pH change.