Question

Question

Data collected for a particular subject during the acid base experiment were

• time between urine collections = 30 min
• urine volume collected = 73 mL
• pH of urine = 7.12
• Buffer-bound H⁺ concentration = 4.5 µmol/mL

Calculate the:

• urinary minute volume (urine flow rate)
• excretion rate of free H⁺ ions
• excretion rate of total H⁺ (free and bound)   

Then compare the excretion rate of free H⁺ ions and that for buffer-bound H⁺ and explain why the H⁺ excretion rate calculated from urine pH underestimates total H⁺ excretion.

Please provide a clear and detailed answer to help me understand. Thank you!

Answer 1

Ans: Data collected were:

Time between urine collections = 30 min, Urine volume collected = 73 mL, pH of urine = 7.12, Buffer-bound H+concentration = 4.5 µmol/mL

(i) Urinary minute volume (urine flow rate) = Urine volume collected / Time between urine collections = 73 mL / 30 min = 2.43 mL/ min

(ii) Excretion rate of free H+ ions:

pH = 7.12, then free [H+] = 0.00000007585 mol/L = 0.00007585 µmol/ mL

So, excretion rate of free H+ ions = (free [H+] x volume) / time = (0.00007585 x 73)/ 30 = 0.000185 µmol/ min

(iii) Excretion rate of total H+ (free and bound) = {(free [H+] +bound [H+]) x volume} / time

= {(0.00007585+4.5) x 73} / 30 = 10.950184 µmol/ min

(iv) Comparison between excretion rate of free [H+] ions and buffer-bound [H+]:

When measuring pH with the pH metre we are only measuring the free [H+] that is excreted out in urine, not taking into account the H+ that bound to the buffer anions (H2PO4). When we add NaOH, the compound will split the H+ from H2PO4- and react with the free H+ to form water. The amount of hydroxide anions added must equal the total free and bound H+ in the urine sample. This is indicated with the use of phenolphthalein, which turns the solution pink when enough NaOH has been added. This will therefore show the most accurate reading of total H+ excretion in urine. Although the H+ excretion rate calculated from simple pH from urine is underestimated, the total free [H+] is so insignificantly small in value (0.000185 µmol/ min) so we can just calculate bound [H+] excretion.