(a) Volume of NaOH
The equation for the reaction is
#"H"_2"C"_2"O"_4 + "2NaOH" → "Na"_2"C"_2"O"_4 + 2"H"_2"O"#
#"Moles of NaOH" = 5.0 × 10^"-3" color(red)(cancel(color(black)("mol H"_2"C"_2"O"_4))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"C"_2"O"_4)))) = "0.0100 mol NaOH"#
#"Volume of NaOH" = 0.0100 color(red)(cancel(color(black)("mol NaOH"))) × "1 L NaOH"/(0.400 color(red)(cancel(color(black)("mol NaOH")))) = "0.0250 L NaOH = 25.0 mL NaOH"#
(b) #K_text(a₁)# for oxalic acid
The equations are:
#"H"_2"C"_2"O"_4 + "H"_2"O" ⇌ "H"_3"O"^"+" + "HC"_2"O"_4^"-"; color(white)(l)K_text(a₁)#
#ul("HC"_2"O"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "C"_2"O"_4^"2-"color(white)(m)); color(white)(l)ul(K_text(a₂))#
#"H"_2"C"_2"O"_4 + "2H"_2"O" ⇌ "2H"_3"O"^"+" + "C"_2"O"_4^"2-"; K#
#K = K_text(a₁)K_text(a₂)#
#K_text(a₁)= K/K_text(a₂) = (3.78 × 10^"-6")/(6.40 × 10^"-5") = 0.0591#
(c) Concentration of oxalate ion
#"pH = 0.5"#
#["H"_3"O"^"+"] = 10^"-0.5"color(white)(l)"mol/L" = "0.32 mol/L"#
We can set up an ICE table for the equilibrium:
#color(white)(mmmmmm)"H"_2"C"_2"O"_4 + "2H"_2"O" ⇌ "2H"_3"O"^"+" + "C"_2"O"_4^"2-"#
#"I/mol·L"^"-1": color(white)(mll)0.015 color(white)(mmmmmmm)0.32color(white)(mmmll)0#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmml)"+"xcolor(white)(mmml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.015-"xcolor(white)(mmmmmll)"0.32+"xcolor(white)(mmlm)x#
#K = (["H"_3"O"^"+"]^2["C"_2"O"_4^"2-"])/(["H"_2"C"_2"O"_4]) = ((0.32+x)^2×x)/(0.015-x) = 3.8 × 10^"-6"#
Check for negligibility
#0.015/(3.8 × 10^"-6") = 4000 > 400#. ∴ #x ≪ 0.015#
Then
#(0.32^2x)/0.015 = 3.8 × 10^"-6"#
#x = (0.015 × 3.8 × 10^"-6")/0.32^2 = 5.7 × 10^"-7"#
#["C"_2"O"_4^"2-"] = x color(white)(l)"mol/L" = 5.7 × 10^"-7"color(white)(l)"mol/L"#
(d) #K_text(b)# for reverse reaction
For the second ionization reaction,
#"HC"_2"O"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "C"_2"O"_4^"2-"; color(white)(ml)K_text(a₂) = 6.40 × 10^"-5" #
For the hydrolysis of oxalate ion,
#"C"_2"O"_4^"2-" + "H"_2"O" ⇌ "HC"_2"O"_4^"-" + "OH"^"-"#
#K_text(b) = K_text(w)/K_text(a₂) = (1.00 × 10^"-14")/(6.40 × 10^"-5") = 1.56 × 10^"-10"#
