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

To calculate the moles of NaOH, we'll use the formula: moles = molarity x volume (in liters). The given molarity of NaOH is 0.200 M, and its volume is 50.00 mL, which is equal to 0.050 L (since 1 L = 1000 mL). So, moles of NaOH = \(0.200 \times 0.050 = 0.010\) moles.

We will now write the balanced chemical equations for the reactions of NaOH with each acid. a. Reaction with HCl: \(NaOH + HCl \rightarrow NaCl + H_{2}O\) b. Reaction with HNO3: \(NaOH + HNO_{3} \rightarrow NaNO_{3} + H_{2}O\) c. Reaction with HC2H3O2: (Since it has 1 acidic hydrogen) \(NaOH + HC_{2}H_{3}O_{2} \rightarrow NaC_{2}H_{3}O_{2} + H_{2}O\) All the reactions have a 1:1 stoichiometry between NaOH and the corresponding acid.

Now, we will use the moles of NaOH and the molarities of each acid to determine their respective volumes for the complete reaction. a. For 0.100 M HCl: Since the reaction has a 1:1 stoichiometry, moles of HCl = moles of NaOH = 0.010 moles. Volume (in liters) = moles / molarity = \(0.010 / 0.100 = 0.100 L\), or \(100.00\ mL\) b. For 0.150 M HNO3: Since the reaction has a 1:1 stoichiometry, moles of HNO3 = moles of NaOH = 0.010 moles. Volume (in liters) = moles / molarity = \(0.010 / 0.150 = 0.0667 L\), or \(66.67\ mL\) c. For 0.200 M HC2H3O2 (1 acidic hydrogen): Since the reaction has a 1:1 stoichiometry, moles of HC2H3O2 = moles of NaOH = 0.010 moles. Volume (in liters) = moles / molarity = \(0.010 / 0.200 = 0.050 L\), or \(50.00\ mL\)

The volumes of each acid required to react completely with 50.00 mL of 0.200 M NaOH are: 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