What is the molar concentration of hydrogen ions if 12 drops of a benzoic acid #HC_7H_5O_2# solution are titrated with 26 drops of a .40 M #KOH# solution?

Question

Mateo Aguilar · Accepted Answer

#["H"^+] = 1.3 × 10^"-9" color(white)(l)"mol/L"#

Warning! Long Answer.

I assume that you have titrated to the end point.

The equation for the reaction is

#"HA" + "KOH" → "KA" +"H"_2"O"#

Since #"HA"# and #"KOH"# react in a 1:1 molar ratio, we can write

#c_aV_a =c_bV_b#

∴ #c_a = c_b × V_b/V_a = "0.40 mol/L" × (26 color(red)(cancel(color(black)("drops"))))/(12 color(red)(cancel(color(black)("drops")))) = "0.867 mol/L"#

At neutralization, you have only #"KA"# present.

If there had been no change in volume, its concentration would have been 0.867 mol/L.

However, the volume increased during the titration to #"(12 drops + 26 drops) = 38 drops"#.

To calculate the new concentration, we can use the dilution formula

#color(blue)(|bar(ul(color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#

#c_1 = "0.867 mol/L"; V_1 = "12 drops"# #c_2 = ?; color(white)(mmmmml)V_2 = "38 drops"#

#c_2 = c_1 × V_1/V_2 = "0.867 mol/L" × (12 color(red)(cancel(color(black)("drops"))))/(28color(red)(cancel(color(black)( "drops")))) = "0.372 mol/L"#

∴ At neutralization, we have a 0.372 mol/L solution of #"KA"#.

Calculate the concentration of #["H"^+]# in a 0.372 mol/L solution of KA

#color(white)(mmmmmmmmll)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"# #"I/mol·L"^"-1": color(white)(mml)0.372color(white)(mmmmmll) 0color(white)(mmll) 0# #"C/mol·L"^"-1":color(white)(mmm) "-"xcolor(white)(mmmmml) +x color(white)(m)+x# #"E/mol·L"^"-1": color(white)(m)0.372 -xcolor(white)(mmmmll) x color(white)(mmll)x#

#K_b = K_w/K_a = (1.00 × 10^"-14")/(6.46 × 10^"-5") = 1.55 × 10^"-10"#

#K_b = (["HA"]["OH"^"-"])/(["A"^"-"])#

#1.55 × 10^"-10" = (x × x)/(0.372 -x) - x^2/(0.372-x)#

Check if #x# is negligible

#0.372/(1.55 × 10^"-10") = 2.11 × 10^9 > 400; ∴ x « 0.372#

#1.55 × 10^"-10" = x^2/0.372#

#x^2 = 0.372 × 1.55 × 10^"-10" = 5.77 × 10^"-11"#

#x = ["OH"^"-"] = 7.59 ×10^"-6"#

#["H"^+] = K_w/["OH"^"-"] = (1.00 × 10^"-14")/(7.59 ×10^"-6") = 1.3 × 10^"-9"color(white)(l) "mol/L"#

Kai Bowman · Answer

undefined To find the molar concentration of hydrogen ions, first calculate the moles of benzoic acid (HC7H5O2) using the volume and molarity of KOH solution. Then, use the stoichiometry of the reaction to determine the moles of hydrogen ions produced. Finally, divide the moles of hydrogen ions by the total volume of the solution in liters to get the molar concentration.