Solve an equilibrium problem (using an ICE table) to calculate the pH of of each solution: a solution that is 0.18M of HCHO2 in 0.14M of NaCHO2. What is the pH?

Oct 24, 2015

$\text{pH} = 3.64$

Explanation:

The first important thing to notice here is that you're actually dealing with a buffer solution that contains formic acid, $\text{CHCOOH}$, a weak cid, and sodium formate, $\text{NaCHCOO}$, a salt of its conjugate base, the formate anion, ${\text{CHCOO}}^{-}$.

Before doing any calculation, you need to know the value of the acid dissociation constnt, ${K}_{a}$, for formic acid, which is listed as $1.8 \cdot {10}^{- 4}$.

Now, the idea here is that the aid will dissociate in aqueous solution to form hydronium ions, ${\text{H"_3"O}}^{+}$, and formate ions.

Since the solution will also contain a significant amount of formate ions delivered by the sodium formate, you will be deling with a common ion effect - think Le Chatelier's Principle.

In other words, the significant concentration of the formate ions will push the equilibrium to the left, so you can expect the formic acid to be less ionized that it would have been if present by itself in aqueous solution.

This implies that you will ahve a smaller concentration of hydronium ions, and in turn a higher pH for this buffer solution.

So, use an ICE table to calculate the equilibrium concentration of the hydronium ions

${\text{CHCO"_2"H"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "CHCO"_text(2(aq])^(-) " "+" " "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

color(purple)("I")" " " "0.18" " " " " " " " " " " " " " " "0.14" " " " " " " " " " " " "0
color(purple)("E")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)
color(purple)("C")" "0.18-x" " " " " " " " " " " " " "0.14+x" " " " " " " " "x

By definition, the acid dissociation constant will be

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["CHCO"_2^(-)])/(["CHCO"_2"H}\right]\right)$

${K}_{a} = \frac{\left(0.14 - x\right) \cdot x}{0.18 - x} = 1.8 \cdot {10}^{- 4}$

Since ${K}_{a}$ is very small, you can use the following approximations

$0.14 - x \approx 0.14 \text{ }$ and $\text{ } 0.18 - x \approx 0.18$

This will give you

$1.8 \cdot {10}^{- 4} = \frac{0.14 x}{0.18} \implies x = \frac{0.18}{0.14} \cdot 1.8 \cdot {10}^{- 4} = 2.3143 \cdot {10}^{- 4}$

This means that the concentration of the hydronium ions will be

["H"_3"O"^(+)] = x = 2.3143 * 10^(-4)"M"

The pH of the solution will thus be

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

$\text{pH} = - \log \left(2.3143 \cdot {10}^{- 4}\right) = \textcolor{g r e e n}{3.64}$

SIDE NOTE You can check your calculations by using the Henderson - Hasselbalch equation

"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))

In your case, that would come out to be

"pH" = - log(K_a) + log ((0.14color(red)(cancel(color(black)("M"))))/(0.18color(red)(cancel(color(black)("M")))))

$\text{pH} = 3.745 + \left(- 0.109\right) = 3.636 \approx 3.64 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$

As practice, try calculating the pH of a 0.18 M formic acid solution and compare it with this one.

If our prediction was correct, it should come out to be smaller than 3.64.