Step 1. Calculate the desired ["H"_3"O"^"+"][H3O+]
["H"_3"O"^"+"] = 10^"-pH"color(white)(l)"mol/L" = 10^"-3.7"color(white)(l)"mol/L" = 2.0 × 10^"-4"color(white)(l)"mol/L"[H3O+]=10-pHlmol/L=10-3.7lmol/L=2.0×10-4lmol/L
Step 2. Calculate the concentration of formic acid
We can use an ICE table to help with the calculation.
color(white)(mmmmmml)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"mmmmmmlHA+H2O⇌H3O++A-
"I/mol·L"^"-1": color(white)(mm)c color(white)(mmmmmmml)0color(white)(mmm)0I/mol⋅L-1:mmcmmmmmmml0mmm0
"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"xC/mol⋅L-1:ml-xmmmmmml+xmml+x
"E/mol·L"^"-1": color(white)(m)c"-"xcolor(white)(mmmmmmm)xcolor(white)(mmm)xE/mol⋅L-1:mc-xmmmmmmmxmmmx
K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(c-x) = 2.0 × 10^"-4"Ka=[H3O+][A-][HA]=x2c−x=2.0×10-4
x^2 = (c-x)(2.0 × 10^"-4") = 2.0 × 10^"-4"c - 2.0 × 10^"-4"xx2=(c−x)(2.0×10-4)=2.0×10-4c−2.0×10-4x
(2.0 × 10^"-4")c = x^2 + 2.0 × 10^"-4"x(2.0×10-4)c=x2+2.0×10-4x
c = (x^2 + 2.0 × 10^"-4"x)/(2.0 ×10^"-4")c=x2+2.0×10-4x2.0×10-4
x = 2.0 × 10^"-4"x=2.0×10-4
∴ c = ((2.0 × 10^"-4")^color(red)(cancel(color(black)(2))) + (2.0 × 10^"-4")^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(2.0 × 10^"-4")))) = 4.0 × 10^"-4"
c = 4.0 × 10^"-4"color(white)(l)"mol/L"
Step 3. Calculate the moles of formic acid
"Moles" = 10 color(red)(cancel(color(black)("L"))) × (4.0 × 10^"-4"color(white)(l)"mol")/(1 color(red)(cancel(color(black)("L")))) = 4.0 × 10^"-3"color(white)(l)"mol"
Step 4. Calculate the mass of formic acid
"Mass" = 4.0 × 10^"-3" color(red)(cancel(color(black)("mol"))) × "46.03 g"/(1 color(red)(cancel(color(black)("mol")))) = "0.18 g"
