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To calculate the initial pH of the acetic acid solution, we first need the dissociation constant (Ka) for acetic acid, which is \(1.8 \times 10^{-5}\). The ionization of acetic acid in water is given by: HC鈧侶鈧僌鈧� (aq) + H鈧侽 (l) 鈬� H鈧僌鈦� (aq) + C鈧侶鈧僌鈧傗伝 (aq) We can use the Ka expression to find the initial concentration of H鈧僌鈦� ions in the solution: Ka = \(\frac{[H_{3}O^{+}] [C_{2}H_{3}O_{2}^{-}]}{[HC_{2}H_{3}O_{2}]}\) Since the solution is only 1.00 M acetic acid, and no NaOH has been added yet, the concentration of C鈧侶鈧僌鈧傗伝 ions will be negligible, and we can approximate the concentration of HC鈧侶鈧僌鈧� as 1.00 M. Therefore, the Ka expression simplifies to: \(1.8 \times 10^{-5}\) = \(\frac{[H_{3}O^{+}]^{2}}{1.00}\) By solving for [H鈧僌鈦篯, we can calculate the initial pH as: pH = -log[H鈧僌鈦篯

After determining the initial pH, we need to find the final pH when it doubles. To do this, we can simply multiply the initial pH by 2: Final pH = 2 脳 Initial pH

Now, we need to find out how many moles of NaOH are required to achieve the final pH. The balanced equation for the reaction between acetic acid and NaOH is: HC鈧侶鈧僌鈧� (aq) + OH鈦� (aq) 鈫� C鈧侶鈧僌鈧傗伝 (aq) + H鈧侽 (l) We can set up an ICE table to assist in calculations: Initial: 1.00 M 0 M Change: -x +x Equilibrium: 1.00 M - x x We can write the expression for the concentration of the conjugate base C鈧侶鈧僌鈧傗伝 at the final pH: [OH鈦籡 = 10^(-14) / [H鈧僌鈦篯 Substitute the value of [H鈧僌鈦篯 from Step 2 in the above expression to find the concentration of OH鈦� ions at the final pH. Now, we can find the moles of NaOH (OH鈦� ions) required to reach the final pH using the one-to-one stoichiometry of the reaction: Moles of NaOH = (Moles of OH鈦�) = (Concentration of OH鈦�) 脳 (Volume of the solution)

To calculate the mass of NaOH to be added, we need to convert the moles of NaOH we found in step 3 to grams by using the molar mass of NaOH: Mass of NaOH = (Moles of NaOH) 脳 (Molar mass of NaOH) The molar mass of NaOH is approximately 40.00 g/mol. Multiply the number of moles by the molar mass to obtain the mass of NaOH in grams. #Solution# Now, we can apply the above steps to find the answer to the given exercise: Step 1: Solve for [H鈧僌鈦篯 and find the initial pH \(1.8 \times 10^{-5}\) = \([H_{3}O^{+}]^{2}\) 鈫� [H鈧僌鈦篯 = \(1.34 \times 10^{-3} M\) pH = -log(1.34 脳 10^(-3)) 鈮� 2.87 Step 2: Calculate the final pH Final pH = 2 脳 2.87 鈮� 5.74 Step 3: Determine the moles of NaOH to be added [H鈧僌鈦篯 at final pH = 10鈦烩伒.鈦封伌 鈮� \(1.83 \times 10^{-6}\) [OH鈦籡 = 10^(-14) / \(1.83 \times 10^{-6}\) 鈮� 5.47 脳 10鈦烩伖 M Moles of NaOH = (5.47 脳 10鈦烩伖 M)(1.00 L) 鈮� 5.47 脳 10鈦烩伖 mol Step 4: Calculate the mass of NaOH Mass of NaOH = (5.47 脳 10鈦烩伖 mol)(40.00 g/mol) 鈮� 2.19 脳 10鈦烩伔 g Therefore, to double the pH of the 1.00 L solution of 1.00 M acetic acid, we need to add approximately \(2.19 \times 10^{-7}\) grams of sodium hydroxide.