Question #3cab0

1 Answer
Apr 7, 2017

Answer:

#["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"#