# pH ChangesWatch

#1
If 0.1mmol/L H+ is added to an unbuffered solution at pH 7.4, the pH of the solution will fall by 3.4 units, but if added to a HCO3- buffered solution the pH falls by just 0.04 units, whilst in the body, as [CO2] remains unchanged at 1.2mmol/L due to an elevated ventilation, the fall in pH will be just 0.002 units.

Use the Henderson-Hasselbalch equation to explain these results.

I really don't understand the answer:
pH = -log [H+] and the HH equation pH = 6.1 +log10 ([HCO3-] / ([CO2] x sol const)) or for normal values 7.4 = 6.1 + log10 (24 / 1.2) (assuming PCO2 = 5kPa and solubility constant is 0.23 mmol L-1 kPa-1)

eg Adding 0.1 mmol/L H+ (ie 0.0001mol/L) to:

1. Unbuffered soln at pH 7.4 (ie 0.00000004 mol/L) would give a final pH of 4 and a ∆pH of 3.4

2. Bicarbonate soln at pH 7.4. The [HCO3-] falls by 0.1mmol/L to 23.9 and the [CO2] rises by 0.1mmol/L to 1.3. Use HH to show that new pH is now 7.36 and so ∆pH is just 0.04 units

3. Bicarbonate soln at pH 7.4 with ventilation keeping PCO2 at 1.2mmol/L (ie 5kPa or 40mmHg). As in 2. but now [CO2] remains at 1.2. Use HH to show that new pH is 7.398 and so ∆pH is only 0.002 units.

How do you work out each answer?

Thanks
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