Calculating [CO2] given pH and [HCO3-]

Moderators: Chem_Mod, Chem_Admin

Leonardo Le Merle 1D
Posts: 56
Joined: Wed Feb 27, 2019 12:16 am

Calculating [CO2] given pH and [HCO3-]

Postby Leonardo Le Merle 1D » Tue Feb 11, 2020 10:23 am

I've been looking at past study guides and one question kind of threw me off, as 10^-6.1 is used for both K and [H3O+]

H3O+(aq) + HCO3− (aq) ⇌ 2H2O(l) + CO2(aq); K = 7.9 × 10-7
This reaction assumes that all H2CO3 produced decomposes completely to CO2 and H2O. Suppose that 1.0 L of blood is brought to pH = 6.1. If the concentration of HCO3- is (5.5 μmol/L), calculate the amount (in moles) of CO2 present in the 1.0 L sample at pH = 6.1. (1 μmol/L-1 = 10-6 mol/L)

Essentially the answer used the given pH to find K=7.94*10^-7, but then also to show that [H3O+] was 7.94*10^-7. Then it plugged the given values into K=[CO2]/[H3O+][HCO3-] to find [CO2].

How is pH used to find both K and [H3O+]?

Jasmine Vallarta 2L
Posts: 102
Joined: Sat Aug 17, 2019 12:18 am

Re: Calculating [CO2] given pH and [HCO3-]

Postby Jasmine Vallarta 2L » Tue Feb 11, 2020 1:45 pm

this is the pH at equilibrium


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests