# 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

###### FREE Expert Solution

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 Ka 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 H(CO2 H) (CO)(aq)

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

Step 2: Write the Ka is an equilibrium expression.

Solids and liquids are not included in the expression

Step 3: Calculate for the equilibrium concentration.

$\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{-}\mathbf{x}\right]}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{=}\frac{{x}^{2}}{\left[0.01-x\right]}$

we cannot ignore x in the denominator

$\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{015}\mathbf{-}\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}\left(\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{-}\mathbf{x}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{0}}{\mathbf{.}}{\mathbf{0011}}{\mathbf{x}}{\mathbf{-}}{\mathbf{0}}{\mathbf{.}}{\mathbf{000011}}{\mathbf{=}}{\mathbf{0}}}$

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

$\overline{){\mathbf{x}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{b}\mathbf{±}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

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 H(CO2 H) (CO)] = [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−(aq)            Ka = 3.9 × 10−6

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

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###### Problem Details

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 H(CO2 H) (CO)(aq)        Ka = 1.1 × 10−3
C6H4(CO2 H)(CO2)(aq) + H2 O(l) ⇌ H3O+(aq) + C6 H4(CO2)2−(aq)            Ka = 3.9 × 10−6