Find the concentration of monomeric dichloroacetic acid in a solution which contains 0.0129 g of the — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Analytical Concentration of the Acid Dichloroacetic acid has the chemical formula \ce{Cl2CHCOOH}. The molar mass of the monomeric form is calculated as: \[M_{\ce{Cl2CHCOOH}} = 2 \times 12 + 2 \times 1 + 2 \times 35.5 + 2 \times 16 = 129\text{ g mol}^{-1}\] Because carboxylic acids in non-polar solvents like carbon tetrachloride (\ce{CCl4}) undergo significant dimerization via intermolecular hydrogen bonding, the acid exists primarily as a dimeric complex, \ce{(Cl2CHCOOH)2}, which has a molar mass of: \[M_{\ce{(Cl2CHCOOH)2}} = 2 \times 129 = 258\text{ g mol}^{-1}\] The analytical concentration of the acid is calculated based on its dimeric formula mass: \[C_{\text{total}} = \frac{\text{Mass}}{\text{Molar Mass} \times \text{Volume}}\] \[C_{\text{total}} = \frac{0.0129\text{ g}}{258\text{ g mol}^{-1} \times 0.1\text{ L}} = 5.0 \times 10^{-4}\text{ M}\] This represents the total analytical concentration of the acid in terms of the monomeric conservation equation: \[[\ce{HA}] + 2[\ce{(HA)2}] = 5.0 \times 10^{-4}\text{ M}\] Step 2 - Define the Equilibrium System and Conservation of Mass The dissociation of the dimeric acid back into its monomeric form in solution is represented as: \[\ce{(HA)2 <=> 2HA}\] Let $x$ be the equilibrium concentration of the monomeric acid, $[\ce{HA}]$: \[[\ce{HA}] = x\text{ M}\] Using the mass conservation relation from Step 1, we find the equilibrium concentration of the dimer, $[\ce{(HA)2}]$: \[[\ce{(HA)2}] = \frac{5.0 \times 10^{-4} - x}{2}\] Step 3 - Set up the Dissociation Constant Expression The equilibrium dissociation constant ($K_d$) for the dimeric acid is expressed as: \[K_d = \frac{[\ce{HA}]^2}{[\ce{(HA)2}]}\] We are given $K_d = 5.0 \times 10^{-4}\text{ M}$. Substituting our expressions into this formula: \[5.0 \times 10^{-4}\text{ M} = \frac{x^2}{\frac{5.0 \times 10^{-4}\text{ M} - x}{2}}\] \[5.0 \times 10^{-4} = \frac{2x^2}{5.0 \times 10^{-4} - x}\] Step 4 - Solve the Quadratic Equation Rearranging the equation into standard quadratic form ($ax^2 + bx + c = 0$): \[2x^2 = 5.0 \times 10^{-4}\left(5.0 \times 10^{-4} - x\right)\] \[2x^2 = 2.5 \times 10^{-7} - 5.0 \times 10^{-4}x\] \[2x^2 + 5.0 \times 10^{-4}x - 2.5 \times 10^{-7} = 0\] We solve for $x$ using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting $a = 2$, $b = 5.0 \times 10^{-4}$, and $c = -2.5 \times 10^{-7}$: \[x = \frac{-5.0 \times 10^{-4} + \sqrt{\left(5.0 \times 10^{-4}\right)^2 - 4(2)\left(-2.5 \times 10^{-7}\right)}}{2(2)}\] \[x = \frac{-5.0 \times 10^{-4} + \sqrt{2.5 \times 10^{-7} + 20 \times 10^{-7}}}{4}\] \[x = \frac{-5.0 \times 10^{-4} + \sqrt{22.5 \times 10^{-7}}}{4}\] Converting $22.5 \times 10^{-7}$ to $2.25 \times 10^{-6}$ to extract the square root: \[\sqrt{2.25 \times 10^{-6}} = 1.5 \times 10^{-3} = 15 \times 10^{-4}\] Substituting back: \[x = \frac{-5.0 \times 10^{-4} + 15 \times 10^{-4}}{4}\] \[x = \frac{10 \times 10^{-4}}{4} = \boxed{2.5 \times 10^{-4}\text{ M}}\] Step 5 - Explain Each Option * **Option (A) $5.0 \times 10^{-4}\text{ M}$**: Incorrect. This represents the total analytical concentration of the acid in the dimeric state or the value of the dissociation constant $K_d$, not the concentration of the monomer. * **Option (B) $2.5 \times 10^{-4}\text{ M}$**: Correct. As mathematically demonstrated, solving the system of equations with the correct mass balance yields exactly this concentration. * **Option (C) $1.0 \times 10^{-3}\text{ M}$**: Incorrect. This is the total concentration of monomer units in the solution if dimerization is completely ignored. * **Option (D) $2.5 \times 10^{-3}\text{ M}$**: Incorrect. This is a scale error.