# What is the "pH" of a "0.20-M" solution of formic acid?

## $p {K}_{a} = 3.75$

Aug 30, 2015

For part (a): $p {H}_{\text{sol}} = 2.22$

#### Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, $\text{HCOOH}$, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

${\text{HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "HCOO"_text((aq])^(-) " "+" " "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

color(purple)("I")" " " " "0.20" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)
color(purple)("E")" "(0.20-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x

You need to use the acid's $p {K}_{a}$ to determine its acid dissociation constant, ${K}_{a}$, which is equal to

$\textcolor{b l u e}{{K}_{a} = {10}^{- p {K}_{a}}}$

${K}_{a} = {10}^{- 3.75} = 1.78 \cdot {10}^{- 4}$

You know that the equilibrium constant is also equal to

${K}_{a} = \left(\left[\text{HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH}\right]\right)$

${K}_{a} = \frac{x \cdot x}{0.20 - x}$

Since ${K}_{a}$ is small compared with the initial concentration of the acid, you can approximate $\left(0.20 - x\right)$ with $0.20$. This will give you

${K}_{a} = {x}^{2} / 0.2 \implies x = \sqrt{0.2 \cdot 1.78 \cdot {10}^{- 4}} = 5.97 \cdot {10}^{- 3}$

The concentration of the hydronium ions will thus be

x = ["H"_3"O"^(+)] = 5.97 * 10^(-3)"M"

This means that the solution's pH will be

pH_"sol" = -log(["H"_3"O"^(+)])

$p {H}_{\text{sol}} = - \log \left(5.97 \cdot {10}^{- 3}\right) = \textcolor{g r e e n}{2.22}$