Problem: Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.C6H4(CO2 H)2(aq) + H2 O(l) ⇌ H3 O+(aq) + C6 H4 (CO2 H) (CO2 )−(aq)        Ka = 1.1 × 10−3C6H4(CO2 H)(CO2)(aq) + H2 O(l) ⇌ H3O+(aq) + C6 H4(CO2)2 2−(aq)            Ka = 3.9 × 10−6

We are being asked to calculate each species present in a 0.010-M solution of phthalic acid, C₆H₄(CO₂H)₂.

Step 1: Construct an ICE chart for the equilibrium reaction.

Since we’re dealing with a weak acid and K_a is an equilibrium expression, we will have to create an ICE chart to determine the equilibrium concentration of each species

We will use the first equilibrium reaction given:

       C₆ H₄(CO₂ H)₂(aq) + H₂ O(l) ⇌ H₃ O⁺(aq) + C₆ H₄ (CO₂ H) (CO₂ )⁻(aq)

        phthalic acid (aq) + H₂ O(l) ⇌ H₃ O⁺(aq) + biphthalate (aq)

Step 2: Write the K_a is an equilibrium expression.

$$\begin{gathered} \boxed{{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}\mathbf{biphthalate}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{[\mathrm{phthalic} \mathrm{acid}]} \end{gathered}$$

Solids and liquids are not included in the expression

Step 3: Calculate for the equilibrium concentration.

$$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{-}\mathbf{x}\right]} \\ \mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{=}\frac{x^2}{[0.01-x]} \end{gathered}$$

$\frac{{\left[\mathbf{phthalic}\mathbf{ }\mathbf{acid}\right]}_{\mathbf{i}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{M}}{\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\mathbf{\ll }\mathbf{500}\mathbf{;}$     we cannot ignore x in the denominator

$$\begin{gathered} \mathbf{(}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{015}\mathbf{-}\mathbf{x}} \\ \mathbf{(}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{)}\left(\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{-}\mathbf{x}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \\ \boxed{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{0011}\mathbf{x}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{000011}\mathbf{=}\mathbf{0}} \end{gathered}$$

Now, we will use the quadratic formula to solve for x. The quadratic formula is:

$\boxed{\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

From the equation, we have a = 1, b = 0.011, and c = -0.000011

Solve for x using the solver function from the calculator:

x = -0.0039  (can't be negative) ; x =0.002812

Hence x = 0.002812 M = [bipthalate] = [C₆ H₄ (CO₂ H) (CO₂ )⁻] = [H₃O⁺]

phtalic acid = 0.010-0.002812 = 0.007188 M

Next is the 2nd equilibrium reaction given:

       C₆H₄(CO₂ H)(CO₂)(aq) + H₂ O(l) ⇌ H₃O⁺(aq) + C₆ H₄(CO₂)₂ ²⁻(aq)            K_a = 3.9 × 10⁻⁶

         biphthalate (aq) + H₂ O(l) ⇌ H₃ O⁺(aq) + phthalate (aq)