Question

Question #3cab0

Answer 1

`["HCO"_3^"-"] = "0.190 mol/L"`

Explanation:

The chemical equation for the buffer is

`"H"_2"CO"_3 + "H"_2"O" ⇌ -"H"_3"O"^"+" + "HCO"_3^"-"`

We can use the Henderson-Hasselbalch equation to calculate the concentration of `"HCO"_3^"-"`.

`"pH" = "p"K_text(a) + log((["HCO"_3^"-"])/(["H"_2"CO"_3]))`

`7.30 = 6.95 + log((["HCO"_3^"-"])/("0.085 mol/L"))`

`log((["HCO"_3^"-"])/("0.085 mol/L")) = 0.35`

`(["HCO"_3^"-"])/("0.085 mol/L") = 10^0.35 = 2.239`

`["HCO"_3^"-"] = 2.239 × "0.085 mol/L" = "0.190 mol/L"`