# Question cb24c

Oct 24, 2015

$\text{pH} = 2.54$

#### Explanation:

Your strategy here is to use an ICE table to find the equilibrium concentration of the hydronium ions, ${\text{H"_3"O}}^{+}$, formed in solution by the partial ionization of formic acid (methanoic acid), $\text{HCO"_2"H}$, a weak acid.

To get the acid dissociation constant, ${K}_{a}$, use the given $p K a$

${K}_{a} = {10}^{- p {K}_{a}}$

${K}_{a} = {10}^{- 3.75} = 0.00017778 = 1.78 \cdot {10}^{- 4}$

So, use an ICE table to get 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")" " " "0.05" " " " " " " " " " " " " " " "0" " " " " " " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " " " " "(+x)
color(purple)("E")" "0.05-x" " " " " " " " " " " " " "x" " " " " " " " " " " " " " "x

By definition, the acid dissociation constant will be

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

${K}_{a} = \frac{x \cdot x}{0.05 - x} = 1.78 \cdot {10}^{- 4}$

Since the initial concentration of the acid is relatively small, you cannot use the approximation

$0.05 - x \approx 0.05$

This means that will have to solve for $x$ by using a quadratic equation

${x}^{2} = 1.78 \cdot {10}^{- 4} \cdot \left(0.05 - x\right)$

${x}^{2} = 8.9 \cdot {10}^{- 6} - 1.78 \cdot {10}^{- 4} x$

${x}^{2} + 1.78 \cdot {10}^{- 4} x - 8.9 \cdot {10}^{- 6} = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since $x$ symbolizes concentration, the only solution that will have chemical and physical significance will be the positive one

$x = 0.0028956$

This means that the concentration of hydronium ions will be

["H"_3"O"^(+)] = x = "0.0028956 M"

The pH of the solution will be

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

$\text{pH} = - \log \left(0.0028956\right) = \textcolor{g r e e n}{2.54}$

SIDE NOTE You can try to solve by using the approximation

$0.05 - x \approx 0.05$

the pH of the solution will be similar, but the approximation error will be greater than 5%, which indicates that the approximation is not justified.

$0.05 - 0.002983 = 0.04702$

The error is

|0.04702 - 0.05|/0.05 xx 100 = 5.96%#