A certain acid, HA, has a vapor density of \(5.11 \mathrm{~g} / \mathrm{L}\) when in the gas phase at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(1.00 \mathrm{~atm}\). When \(1.50 \mathrm{~g}\) of this acid is dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution, the \(\mathrm{pH}\) is found to be \(1.80\). Calculate \(K_{\mathrm{a}}\) for the acid.

To find the molar mass of the acid, we will use the ideal gas law equation, PV = nRT. We will first rearrange the equation to find n (number of moles), and then we will use the vapor density to find the molar mass of the acid. Rearrange the ideal gas law equation: n = PV/RT Given: Vapor density = \(5.11 \frac{g}{L}\) Temperature (T) = \(25^{\circ}\) C = 298.15 K (converted to Kelvin) Pressure (P) = 1.00 atm R (gas constant) = 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\) Volume (V) = 1 L (assuming 1 L of gas) Now we plug in the values into the equation: n = \(\frac{(1.00 \, atm)(1 \, L)}{(0.0821 \, \frac{L \cdot atm}{mol \cdot K})(298.15 \, K)}\) n = 0.04094 mol Now, we use the vapor density to find the molar mass of the acid: Molar mass = \(\frac{mass}{moles}\) Molar mass = \(\frac{5.11 \,\frac{g}{L}}{0.04094 \, \frac{mol}{L}}\) Molar mass = 124.7 g/mol

Next, we will find the concentration of the acid, [HA], in the 100.0 mL solution. We are given the mass of HA dissolved in the solution (1.50 g) and its molar mass (124.7 g/mol) from Step 1. Concentration = \(\frac{moles}{volume}\) First, convert mass to moles of HA: moles = \(\frac{1.50 \, g}{124.7 \,\frac{g}{mol}}\) = 0.01203 mol Now, convert mL to L and find the concentration: [HA] = \(\frac{0.01203 \, mol}{0.100 \, L}\) = 0.1203 M

We are given the pH of the solution, which is 1.80. We will use this to find the concentration of H+ ions in the solution. [H+] = \(10^{-pH}\) [H+] = \(10^{-1.80}\) = 0.0158 M Since HA is a weak acid, it does not fully dissociate in water. However, we can safely assume that the concentration of [A-] ions is approximately equal to the concentration of [H+] ions because for each HA molecule that dissociates, one H+ and one A- ion are produced. [A-] = 0.0158 M Now that we know the initial concentration of HA and the concentrations of H+ and A- ions after dissociation, we can update [HA]: [HA] (updated) = 0.1203 M - 0.0158 M = 0.1045 M

Finally, we can use the concentrations of HA, H+, and A- ions to calculate the Ka (acid dissociation constant) of the acid. Ka = \(\frac{[H^+][A^-]}{[HA]}\) Ka = \(\frac{(0.0158 \, M)(0.0158 \, M)}{0.1045 \, M}\) Ka = 2.39 x 10^-3 Therefore, the Ka of the acid HA is 2.39 x 10^-3.