Calculate the mass of sodium hydroxide that must be added to 1.00 \(\mathrm{L}\) of \(1.00-M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

To calculate the initial pH of the acetic acid solution, we first need to determine the Ka (acid dissociation constant) for acetic acid. The Ka of acetic acid is 1.8 x 10⁻⁵. Then, we can write the equilibrium expression for acetic acid's ionization: \[HC_{2}H_{3}O_{2} \rightleftharpoons H^{+} + C_{2}H_{3}O_{2}^{-}\] Since the initial concentration of acetic acid is given as 1.00 M, we can set up the Ka expression: \[Ka = \frac{[H^{+}][C_{2}H_{3}O_{2}^{-}]}{[HC_{2}H_{3}O_{2}]}\] Plugging in the given values, we get: \[1.8 \times 10^{-5} = \frac{(x)(x)}{1 - x}\] Assuming that x is very small compared to 1, we can ignore the x in the denominator, so we get: \[1.8 \times 10^{-5} = x^2\] Now, we will solve for x (which represents the [H⁺] ion concentration): \[x = \sqrt{1.8\times10^{-5}}\]

Now that we have found the [H⁺] ion concentration, we can use the pH formula to calculate the pH of the acetic acid solution: \[ pH = -\log{[H^{+}]} \] Plug the calculated [H⁺] into the pH formula to find the initial pH: \[ pH = -\log{\sqrt{1.8\times10^{-5}}} \] Once we have the initial pH, we can double it to find the new pH: \[ new\ pH = 2 \times initial\ pH \]

Now that we have the new pH, we can use it to determine the concentration of [H⁺] ions: \[ [H^{+}] = 10^{-new\ pH} \] As NaOH neutralizes acetic acid on a 1:1 basis, the moles of NaOH required to neutralize the moles of [H⁺] ions can be determined directly from the concentration change. \[moles\ of\ NaOH = moles\ of\ H^{+}_{final} - moles\ of\ H^{+}_{initial}\]

Now that we have the moles of NaOH needed, we can convert it to mass by using the molar mass of NaOH. The molar mass of NaOH is 22.99 g/mol (Na) + 15.99 g/mol (O) + 1.01 g/mol (H) = 40 g/mol: \[mass\ of\ NaOH = moles\ of\ NaOH \times molar\ mass\ of\ NaOH\] This will give us the mass of NaOH that must be added to the 1.00 L solution of 1.00 M acetic acid to double the pH of the solution.