A \(225-\mathrm{mg}\) sample of a diprotic acid is dissolved in enough water to make \(250 . \mathrm{mL}\) of solution. The \(\mathrm{pH}\) of this solution is \(2.06\). A \(6.9 \times 10^{-3} M\) solution of calcium hydroxide is prepared. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The \(\mathrm{pH}\) at the second equivalence point (as determined by a pH meter) is \(7.96 .\) The first dissociation constant for the acid \(\left(K_{\Delta}\right)\) is \(5.90 \times\) \(10^{-2}\). Assume that the volumes of the solutions are additive, that all solutions are at \(25^{\circ} \mathrm{C}\), and that \(K_{a_{1}}\) is at least 1000 times greater than \(K_{a_{2}}\). a. Calculate the molar mass of the acid. b. Calculate the second dissociation constant for the acid \(\left(K_{\mathrm{a}}\right)\).

Step by step solution

01

Calculate the concentration of the acid solution

We can find the concentration of the acid by dividing the mass by the volume of the solution. We are given that a \(225\,\mathrm{mg}\) sample of the diprotic acid is dissolved in \(250\,\mathrm{mL}\) of water. To find the molar concentration of the acid, we need to convert the mass and volume units to moles per liter (M): \[C_{\mathrm{acid}} = \frac{225\,\mathrm{mg}\times 10^{-3} \mathrm{g/mg}}{250\,\mathrm{mL}\times 10^{-3}\,\mathrm{L/mL}}\]

02

Find the moles of acid per liter

We can calculate the moles of acid per liter by using the formula \[\mathrm{moles}\,\mathrm{acid} = C_{\mathrm{acid}}\times V_{\mathrm{solution}}\] where \(V_{\mathrm{solution}}\) is the volume of the solution in liters and \(C_{\mathrm{acid}}\) is the concentration from step 1.

03

Find the molar mass of the acid using the first dissociation constant

Since the first dissociation constant, \(K_{\Delta}\), is given to be \(5.90\times10^{-2}\), we can express it in terms of the concentrations of hydrogen ions, the acid, and the resulting conjugate base: \[K_{\Delta} = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\] We know that the pH of the solution is \(2.06\), so we can find the concentration of hydrogen ions as follows: \[[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-2.06}\] Using this concentration of hydrogen ions and the given \(K_{\Delta}\) value, we can represent the molar concentration ratio of the acid and conjugate base and solve for the molar mass of the acid.

04

Find the volume of calcium hydroxide solution needed to reach the second equivalence point

We are given that the concentration of the calcium hydroxide solution is \(6.9\times 10^{-3}\,\mathrm{M}\), and we are also given that the pH at the second equivalence point is \(7.96\). To find the volume of the calcium hydroxide solution needed to reach the second equivalence point, we can use the fact that the moles of acid required for the second equivalence point are equal to the moles of the second conjugate base: \[\mathrm{moles}\,\mathrm{acid} = [\mathrm{A^2-}]\times V_{\mathrm{solution}}\] where \([\mathrm{A^2-}]\) is the concentration of the second conjugate base and \(V_{\mathrm{solution}}\) is the volume of the solution. We can then use the \(K_{a_{2}}\) value to solve for the concentration of \([\mathrm{A^2-}]\).

05

Calculate the second dissociation constant using the second equivalence point and the pH

Now that we have the molar mass of the acid and the concentration of the second conjugate base, we can calculate the second dissociation constant using the relationship: \[K_{\mathrm{a}}= \frac{[\mathrm{H^+}][\mathrm{A^2-}]}{[\mathrm{A^-}]}\] We know the pH at the second equivalence point is \(7.96\), so we can find the concentration of hydrogen ions as follows: \[[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-7.96}\] Using this concentration of hydrogen ions and the concentration of the second conjugate base from step 4, we can solve for the second dissociation constant.

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