What volume of each of the following bases will react completely with 25.00 mL of 0.200 M HCl? a. 0.100\(M \mathrm{NaOH}\) b. 0.0500\(M \mathrm{Sr}(\mathrm{OH})_{2}\) c. 0.250 \(\mathrm{M} \mathrm{KOH}\)

We know the volume (25 mL) and molarity (0.200M) of the given HCl solution, so we can calculate the moles of HCl using the following formula: Moles of HCl = Volume (L) × Molarity (M) First, convert 25 mL to liters: \(25\,\mathrm{mL}\times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}=0.025\,\mathrm{L}\) Now, calculate the moles of HCl: Moles of HCl = 0.025 L × 0.200 M = 0.005 mol

For each base, we will determine the number of moles required to completely react with the given moles of HCl. (a) NaOH: HCl + NaOH → NaCl + H₂O (1 mol of HCl reacts with 1 mol of NaOH) Moles of NaOH = 0.005 mol (b) Sr(OH)₂: 2HCl + Sr(OH)₂ → SrCl₂ + 2H₂O (2 mol of HCl react with 1 mol of Sr(OH)₂) Moles of Sr(OH)₂ = 0.005 mol HCl × (1 mol Sr(OH)₂ / 2 mol HCl) = 0.0025 mol (c) KOH: HCl + KOH → KCl + H₂O (1 mol of HCl reacts with 1 mol of KOH) Moles of KOH = 0.005 mol

We will use the moles of each base and their given molarity to calculate the volume required for each base. (a) Volume of 0.100 M NaOH: Volume (L) = Moles of NaOH / Molarity of NaOH = 0.005 mol / 0.100 M = 0.050 L Convert to mL: 0.050 L × 1000 = 50.00 mL (b) Volume of 0.0500 M Sr(OH)₂: Volume (L) = Moles of Sr(OH)₂ / Molarity of Sr(OH)₂ = 0.0025 mol / 0.0500 M = 0.050 L Convert to mL: 0.050 L × 1000 = 50.00 mL (c) Volume of 0.250 M KOH: Volume (L) = Moles of KOH / Molarity of KOH = 0.005 mol / 0.250 M = 0.020 L Convert to mL: 0.020 L × 1000 = 20.00 mL The volume of each base that will react completely with 25 mL of 0.200 M HCl is: a. 50.00 mL of 0.100 M NaOH b. 50.00 mL of 0.0500 M Sr(OH)₂ c. 20.00 mL of 0.250 M KOH