Answer:
Thus, the Quantity of [tex]HCo_{3}^{-}[/tex] Present in urine sample in 24 hours is 35mmol
Explanation:
Acid Production in 12 hours = 40 meq
For, 24 hours = 40 meq × 2
Acid Production in 24 hours = 80 meq
Urine sample for [tex]NH4^{+}[/tex] = 30mmol
[tex]HCo_{3}^{-}[/tex] Present in Urine Sample in 24 hours
= Acid Production - Urine sample for [tex]NH4^{+}[/tex] - Titratable Acid
= 80 - 30 - 15
= 50 - 15
= 35mmol
[tex]HCo_{3}^{-}[/tex] Present in urine sample in 24 hours is 35mmol
