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 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 determine the molar mass of the acid, we can use the ideal gas law. The ideal gas law equation is: \[PV = nRT\] where P is the pressure, V is the volume, n is the amount of substance (in moles), R is the ideal gas constant, and T is the temperature in Kelvin. We are given the vapor density of the gas (\(\rho\)), which is the mass per unit volume (\(\rho = \frac{m}{V}\)). We can rearrange the ideal gas law to solve for the molar mass (\(M = \frac{m}{n}\)). \[PV = \frac{m}{M}RT\] Rearrange to solve for molar mass: \[M = \frac{mRT}{PV}\] Molar mass can be calculated as follows: - Convert the temperature from Celsius to Kelvin: \(T = 25^{\circ}C + 273.15 = 298.15\,\text{K}\) - R (ideal gas constant) = 0.0821 L atm mol⁻¹ K⁻¹ - P (pressure) = 1.00 atm - mass (m) = 5.11 g/L × V - volume (V) = 1 L (to simplify the calculation) Plug in values and solve for the molar mass: \[M = \frac{(5.11\,\text{g/L} \times 1\,\text{L}) \times (0.0821\,\text{L}\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1}) \times (298.15\,\text{K})}{(1.00\,\text{atm}) \times (1\,\text{L})}\] \[M = 126.97\,\text{g/mol}\]

Given that 1.50 g of HA is dissolved in enough water to make 100.0 mL of solution, we can calculate the concentration (in moles per liter) as follows: \[C = \frac{n}{V}\] where C is the concentration, n is the amount of substance in moles, and V is the volume in liters. Since we now have the molar mass of HA, we can convert the mass of HA into moles: \[n = \frac{1.50\,\text{g}}{126.97\,\text{g/mol}} = 0.01182\,\text{mol}\] Now, convert the volume from milliliters to liters: \[V = \frac{100.0\,\text{mL}}{1000\,\text{mL/L}} = 0.100\,\text{L}\] Now we get: \[C_{HA} = \frac{0.01182\,\text{mol}}{0.100\,\text{L}} = 0.1182\,\text{M}\]

The pH of the solution is given as 1.80. Using the definition of pH, we can calculate the concentration of H+ ions: \[\text{pH} = -\log{[H^+]}\] Rearrange and solve for [H+]: \[[H^+] = 10^{-(\text{pH})} = 10^{-1.8} = 0.01585\,\text{M}\] Now, we need to calculate the A- ions concentration. Since the dissociation reaction of the acid is given by: \[HA \rightleftharpoons H^+ + A^-\] This means that when HA loses one H+ ion, it forms one A- ion. Therefore, [A-] = [H+] = 0.01585 M.

Now we have the initial concentration of HA and the concentrations of H+ and A- ions. The equilibrium equation for the dissociation of the acid is: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Using the calculated concentrations: \[K_a = \frac{(0.01585\,\text{M})(0.01585\,\text{M})}{(0.1182\,\text{M} - 0.01585\,\text{M})} = \frac{0.0002515}{0.10235} = 0.002456\] Thus, the acid dissociation constant (\(K_a\)) for the acid is 0.002456.