What is the ratio of hco−3 to h2co3 in an exhausted marathon runner whose blood ph is 7.2?

The mentioned question can be solved with the help of Henderson-Hasselbalch equation. In biochemistry and chemistry, the Henderson-Hasselbalch equation can be used to find out the pH of a buffer solution comprising given concentrations of an acid and its conjugate base.

By using Henderson-Hasselbalch equation:

pH = pKa + log ([HCO³⁻] / [H₂CO₃])

pH = -log Ka + log ([HCO³⁻] / [H₂CO₃])

7.2 = -log (4.3 × 10⁻⁷) + log ([HCO³⁻] / [H₂CO₃])

log ([HCO³⁻] / [H₂CO₃]) = 0.83347

[HCO³⁻] / [H₂CO₃] = 10^0.83 = 6.8

Hence, the ratio of HCO³⁻ to H₂CO₃ in an exhausted runner would be 6.8 to 1.

false

because it does not have the potential to be entirely used up

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g