What volume of each of the following acids will react completely with \(50.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH} ?\) a. \(0.100 \mathrm{M} \mathrm{HCl}\) b. \(0.150 \mathrm{MHNO}_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) ( 1 acidic hydrogen)

Given the volume and concentration of NaOH, we can calculate the moles of NaOH using the formula: Moles of NaOH = Volume (L) × Concentration (M) Since the volume is given in mL, we'll convert it to L by dividing by 1000: Volume in L = 50.00 mL / 1000 = 0.05000 L Now, we can find the moles of NaOH: Moles of NaOH = 0.05000 L × 0.200 M = 0.01000 mol

We'll use stoichiometry to determine the volume of each acid required for a complete reaction with NaOH. The balanced chemical equation for the reaction between each acid and NaOH is: Acid + NaOH → Na+ (aq) + other products The mole ratio between each acid and NaOH is 1:1. Therefore, the moles of each acid required for complete reaction are equal to the moles of NaOH. a. For 0.100 M HCl: Volume of HCl = Moles of HCl / Concentration of HCl = 0.01000 mol / 0.100 M = 0.100 L or 100.00 mL b. For 0.150 M HNO3: Volume of HNO3 = Moles of HNO3 / Concentration of HNO3 = 0.01000 mol / 0.150 M = 0.06667 L or 66.67 mL c. For 0.200 M HC2H3O2 (1 acidic hydrogen): Volume of HC2H3O2 = Moles of HC2H3O2 / Concentration of HC2H3O2 = 0.01000 mol / 0.200 M = 0.05000 L or 50.00 mL So, the volume of each acid required for a complete reaction with 50.00 mL of 0.200 M NaOH is: a. 100.00 mL of 0.100 M HCl b. 66.67 mL of 0.150 M HNO3 c. 50.00 mL of 0.200 M HC2H3O2