91影视

First, we need to determine the molar masses of both compounds: For \(\mathrm{KH}_{2} \mathrm{PO}_{4}:\) \(\mathrm{K = 39.10 ~g/mol, ~H_{2} = 2.02 ~g/mol, ~P = 30.97 ~g/mol, ~O_{4} = 64.00 ~g/mol}\) So, molar mass of \(\mathrm{KH}_{2} \mathrm{PO}_{4} = 39.10 + 2.02 + 30.97 + 64.00 = 136.09 \mathrm{~g/mol}\) For \(\mathrm{K}_{2}\mathrm{HPO}_{4}:\) \(\mathrm{K_{2} = 78.20 ~g/mol, ~P = 30.97 ~g/mol, ~O_{4} = 64.00 ~g/mol}\) So, molar mass of \(\mathrm{K}_{2}\mathrm{HPO}_{4} = 78.20 + 30.97 + 64.00 = 173.17 \mathrm{~g/mol}\) Now, we can find the moles of each compound: For \(\mathrm{KH}_{2} \mathrm{PO}_{4}:\) moles = mass / molar mass = \(10.0 \mathrm{~g} / 136.09 \mathrm{~g/mol} = 0.0735 \mathrm{~mol}\) For \(\mathrm{K}_{2}\mathrm{HPO}_{4}:\) moles = mass / molar mass = \(10.0 \mathrm{~g} / 173.17 \mathrm{~g/mol} = 0.0577 \mathrm{~mol}\) Now, calculate the concentrations: For \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}:\) concentration = moles / volume = \(0.0735 \mathrm{~mol} / 0.500 \mathrm{~L} = 0.147 \mathrm{~M}\) For \(\mathrm{HPO}_{4}^{2-}:\) concentration = moles / volume = \(0.0577 \mathrm{~mol} / 0.500 \mathrm{~L} = 0.115 \mathrm{~M}\)

Since \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) dissociates into \(\mathrm{K}^{+}\) and \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), the concentration of \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) is the same as the concentration of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\). For \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}:\) concentration = \(0.147 \mathrm{~M}\) Similarly, since \(\mathrm{K}_{2}\mathrm{HPO}_{4}\) dissociates into T\(2\mathrm{K}^{+}\) and \(\mathrm{HPO}_{4}^{2-}\), the concentration of \(\mathrm{HPO}_{4}^{2-}\) is the same as the concentration of \(\mathrm{K}_{2}\mathrm{HPO}_{4}\). For \(\mathrm{HPO}_{4}^{2-}:\) concentration = \(0.115 \mathrm{~M}\)

We are given that \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) are conjugate acid-base pairs, and we can find the dissociation constant \(K_{a1}\) of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) from a reference: \(K_{a1} (\mathrm{H}_{2}\mathrm{PO}_{4}^{-}) = 6.2 \times 10^{-8}\) To find the \(K_{a}\) for the reaction \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}\), we use the second dissociation constant, \(K_{a2}\) of \(\mathrm{H}_{3}\mathrm{PO}_{4}\): \(K_{a2} (\mathrm{H}_{3}\mathrm{PO}_{4}) = 6.2 \times 10^{-8} / 7.1 \times 10^{-13} = 7.3 \times 10^{-2}\)

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa and the ratio of the concentrations of the conjugate acid \(\mathrm{A}^{-}\) and the conjugate base \(\mathrm{HA}\): \(\mathrm{pH} = \mathrm{p}K_{a} + \log \left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\) For our buffer system, the conjugate acid is \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), and the conjugate base is \(\mathrm{HPO}_{4}^{2-}\). The \(K_{a}\) value for \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) is \(7.3 \times 10^{-2}\). Substitute the values into the equation: pH = \(- \log{(7.3 \times 10^{-2})} + \log \left(\frac{0.115}{0.147}\right)\)

Using a calculator to perform the calculations: pH = \(1.14 - (-0.386) = 1.526\) The pH of the soft drink is approximately 1.53.