How many milliliters of 0.0850\(M \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) 40.0 \(\mathrm{mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3},\) (\mathbf{b} ) 35.0 \(\mathrm{mL}\) of \(0.0850 M \mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) 50.0 \mathrm{mL}\) of a solution that contains 1.85 \(\mathrm{g}\) of \(\mathrm{HCl}\) per liter?

For each acidic solution, we need to find the moles of the solute. We will use the formula: Moles = Molarity × Volume The volume should be in liters for it to be consistent with the unit of molarity. (a) For 40.0 mL of 0.0900 M HNO3, calculate moles of HNO3: Moles of HNO3 = 0.0900 M × (40.0 mL × (1 L / 1000 mL)) Moles of HNO3 = 0.00360 mol (b) For 35.0 mL of 0.0850 M CH3COOH, calculate moles of CH3COOH: Moles of CH3COOH = 0.0850 M × (35.0 mL × (1 L / 1000 mL)) Moles of CH3COOH = 0.002975 mol (c) For 50.0 mL of HCl containing 1.85 g of HCl per liter, first calculate the molarity of HCl: Molarity of HCl = (1.85 g / (36.461 g/mol)) / 1 L Molarity of HCl = 0.05076 M Now, calculate moles of HCl: Moles of HCl = 0.05076 M × (50.0 mL × (1 L / 1000 mL)) Moles of HCl = 0.002538 mol

Now, we will write down the balanced chemical equations for the titrations with NaOH: (a) HNO3 + NaOH → NaNO3 + H2O (b) CH3COOH + NaOH → CH3COONa + H2O (c) HCl + NaOH → NaCl + H2O In each case, the stoichiometric coefficients for both the acid and the base are equal to 1. This means that the number of moles of NaOH required to neutralize the acid completely is equal to the number of moles of the acid.

Now, we will calculate the volume of 0.0850 M NaOH required to neutralize the acid by using the formula: Volume = Moles / Molarity (a) For 0.00360 mol of HNO3: Volume of NaOH = 0.00360 mol / 0.0850 M Volume of NaOH = 0.04235 L (42.35 mL) (b) For 0.002975 mol of CH3COOH: Volume of NaOH = 0.002975 mol / 0.0850 M Volume of NaOH = 0.03500 L (35.00 mL) (c) For 0.002538 mol of HCl: Volume of NaOH = 0.002538 mol / 0.0850 M Volume of NaOH = 0.02986 L (29.86 mL) So, the required volumes of 0.0850 M NaOH to titrate each acidic solution to the equivalence point are: (a) 42.35 mL for 0.0900 M HNO3 (b) 35.00 mL for 0.0850 M CH3COOH (c) 29.86 mL for the HCl solution containing 1.85 g of HCl per liter