What concentration of formic acid will result in a solution with #"pH" = 1.90#? The value of #K_a# for formic acid is #1.8 * 10^(−4)#

The idea here is that the #"pH"# of the solution will give you the equilibrium concentration of hydronium cations in this formic acid solution.

Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use #"HA"# to denote the formic acid and #"A"^(-)# to denote the formate anions.

#"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A"_ ((aq))^(-)#

As you know, the #"pH"# of the solution is given by

#"pH" = - log(["H"_3"O"^(+)])#

This implies that the concentration of hydronium cations is equal to

#["H"_3"O"^(+)] = 10^(-"pH")color(white)(.)"M"#

Now, notice that in order for the ionization of the weak acid to produce #1# mole of hydronium cations and #1# mole of conjugate base, #1# mole of formic acid must ionize.

If you take #["HA"]# to be the equilibrium concentration of formic acid, you can say that the initial concentration of the acid is equal to

#["HA"]_ 0 = ["HA"] + ["H"_ 3"O"^(+)]#

This is equivalent to saying that in order to get an equilibrium concentration of #["H"_3"O"^(+)]# in the solution, the initial concentration of the acid must decrease by #["H"_3"O"^(+)]#.

#["HA"] = ["HA"]_ 0 - ["H"_ 3"O"^(+)]#

Now, by definition, the acid dissociation constant is equal to

#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#

Since you know that, at equilibrium, you have

#["H"_3"O"^(+)] = ["A"^(-)] = 10^(-"pH")#

you can rewrite the expression you have for the acid dissociation constant as

#K_a = (["H"_3"O"^(+)]^2)/(["HA"]_0 - ["H"_3"O"^(+)])#

which is, of course, equivalent to

#K_a = (10^(-"pH"))^2/(["HA"]_0 - 10^(-"pH"))#

Rearrange to solve for the initial concentration of the formic acid

#["HA"]_0 * K_a = 10^(-2"pH") + 10^(-"pH") * K_a#

#["HA"]_0 = 10^(-2"pH")/K_a + 10^(-"pH")#

Finally, plug in your values to get

#["HA"]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)("0.89 M")))#

The answer is rounded to two sig figs, the number of decimal places you have for the #"pH"# of the solution.

Step 1. Calculate #["H"_3"O"^"+"]#

#["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-1.9"color(white)(l)"mol/L" = "0.0126 mol/L"#

Step 2. Calculate the concentration of formic acid

We can use an ICE table to organize our calculations.

Let #c# be the initial concentration of the acid.

#color(white)(mmmmmmll)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
#"I/mol·L"^"-1":color(white)(mml)c color(white)(mmmmmmm)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(m)c-xcolor(white)(mmmmmm)xcolor(white)(mmm)x#

In this problem, #x = "0.015 85"#.

So, our ICE table becomes

#color(white)(mmmmmmmmll)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" color(white)(ll)+color(white)(mll) "A"^"-"#
#"I/mol·L"^"-1":color(white)(mmmm)c color(white)(mmmmmmml)0color(white)(mmmmml)0#
#"C/mol·L"^"-1":color(white)(llm)"-0.0126"color(white)(mmmml)"+0.0126"color(white)(mm)"+0.0126"#
#"E/mol·L"^"-1":color(white)(m)c-"0.0126"color(white)(mmmml)"0.0126"color(white)(mmm)"0.0126"#

#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = ("0.0126 × 0.0126")/(c-"0.0126") = 1.8 × 10^"-4"#

#1.585 × 10^"-4" = 1.8 × 10^"-4"c - 2.268 × 10^"-6"#

#158.5 = 180c +2.268#

#x =(158.5+2.268)/180 = 0.89#

#["HA"] = "0.89 mol/L"#

Check:

#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = ("0.0126 × 0.0126")/(0.89-0.0126) = 1.8 × 10^"-4"#

It checks!