How many milliliters of \(0.0750 \mathrm{M} \mathrm{KOH}\) are required to titrate each of the following solutions to the equivalence point: \((\mathbf{a}) 30.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HCOOH},\) (b) \(45.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{HNO}_{3},\) (c) $50.0 \mathrm{~mL}\( of a solution that contains \)3.00 \mathrm{~g}\( of \)\mathrm{HBr}$ per liter?

Determine the volume of 0.0750 M KOH solution required to titrate 30.0 mL of 0.0900 M HCOOH solution. 1. Write the balanced chemical equation: \(HCOOH + KOH \rightarrow KCOOH + H_2O\) 2. Calculate the moles of \(HCOOH\): moles = concentration × volume moles = \(0.0900 M \times 0.0300\: L\) moles = \(2.70 \times 10^{-3} mol\) 3. Determine the mole ratio between \(HCOOH\) and \(KOH\): From the balanced chemical equation, for each mole of \(HCOOH\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 4. Calculate the volume of the KOH solution required: moles of KOH = moles of \(HCOOH\) moles of KOH = \(2.70 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{2.70 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0360\: L\) or \(36.0\: mL\)

Determine the volume of 0.0750 M KOH solution required to titrate 45.0 mL of 0.0750 M HNO3 solution. 1. Write the balanced chemical equation: \(HNO_{3} + KOH \rightarrow KNO_{3} + H_{2}O\) 2. Calculate the moles of \(HNO_{3}\): moles = concentration × volume moles = \(0.0750 M \times 0.0450\: L\) moles = \(3.375 \times 10^{-3} mol\) 3. Determine the mole ratio between \(HNO_{3}\) and \(KOH\): From the balanced chemical equation, for each mole of \(HNO_{3}\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 4. Calculate the volume of the KOH solution required: moles of KOH = moles of \(HNO_{3}\) moles of KOH = \(3.375 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{3.375 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0450\: L\) or \(45.0\: mL\)

Determine the volume of 0.0750 M KOH solution required to titrate 50.0 mL of a solution containing 3.00 g of HBr per liter. 1. Write the balanced chemical equation: \(HBr + KOH \rightarrow KBr + H_2O\) 2. Calculate the concentration of the HBr solution: Concentration = \(\frac{3.00\: g/L}{80.91\: g/mol}\) (molar mass of HBr = 80.91 g/mol) Concentration = \(0.0371\: M\) 3. Calculate the moles of \(HBr\): moles = concentration × volume moles = \(0.0371\: M \times 0.0500\: L\) moles = \(1.855 \times 10^{-3} mol\) 4. Determine the mole ratio between \(HBr\) and \(KOH\): From the balanced chemical equation, for each mole of \(HBr\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 5. Calculate the volume of the KOH solution required: moles of KOH = moles of HBr moles of KOH = \(1.855 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{1.855 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0247\: L\) or \(24.7\: mL\)