# The venom of biting ants contains formic acid, #"HCOOH"#. What is the #"pH"# of a #"0.0700-M"# solution of formic acid?

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As the feedback and solution both indicate, the difference in calculated #"pH"# when solving with the quadratic equation and the assumption that #x# is small enough differ by only #0.01# #"pH"# units, so using the #x# -is-small assumption is recommended.

#K_a = 1.8 * 10^(–4)# at #25^@"C"# .

As the feedback and solution both indicate, the difference in calculated

##### 1 Answer

#### Explanation:

Formic acid is a **weak acid**, so you know that it will only partially ionize in aqueous solution to produce formate anions and hydronium cations.

#"HCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

Notice that every **mole** of formic acid **that ionizes** produces **mole** of formate anions and **mole** of hydronium cations.

Now, let's say that **that ionizes**. Since formic acid is a weak acid, you know for a fact that

#x quad "M" < "0.0700 M"#

In other words, only a fraction of the initial concentration of the acid will actually ionize. According to the balanced chemical equation, the solution will contain

#["HCOO"^(-)] = ["H"_ 3"O"^(+)] = x quad "M"# When

#x# #"M"# of formic acidionizes, you get#x# #"M"# of formate anions and#x# #"M"# of hydronium cations.

So **at equilibrium**, you can express the concentration of the formate anions and the concentration of the hydronium cations in terms of the concentration of formic acid **that ionizes**. At equilibrium, the resulting solution will also contain

#["HCOOH"] = (0.0700 - x) quad "M"# When

#x# #"M"# of formic acidionizes, the initial concentration of the aciddecreasesby#x# #"M"# .

By definition, the **acid dissociation constant** that describes the ionization equilibrium of formic acid is equal to

#K_a = (["HCOO"^(-)] * ["H"_ 3"O"^(+)])/(["HCOOH"])#

In your case, this will be equal to

#K_a = (x * x)/(0.0700 - x)#

which is

#1.8 * 10^(-4) = x^2/(0.0700 - x)#

Now, you know that using the approximation

#0.0700 - x ~~ 0.0700#

is recommended here, so rewrite the expression of the acid dissociation constant as

# 1.8 * 10^(-4) = x^2/0.0700#

This will get you

#x = sqrt(0.0700 * 1.8 * 10^(-4)) = 0.003550#

Since **equilibrium concentration** of hydronium cations, you can say that the solution will have

#["H"- 3"O"^(+)] = "0.003550 M"#

As you know, the

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

Plug in your value to find

#"pH" = - log(0.003550) = color(darkgreen)(ul(color(black)(2.450)))#

The answer is rounded to three **decimal places** because you have three **sig figs** for the initial concentration of formic acid.