Question #3cab0

Question

1999 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-07T21:06:36

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

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 $"HCO"_3^"-"$ .

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

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

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

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

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