(a) What volume of \(0.115 \mathrm{MHClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{MNaOH}\) ? (b) What volume of $0.128 \mathrm{MHCl}\( is needed to neutralize \)2.87 \mathrm{~g}$ of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\) ), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of a 0.108 \(M\) HCl solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of KOH must be present in the solution?

To find the volume of the HClO4 solution needed, we will use the balanced chemical equation and the concept of moles: \[HClO_{4} + NaOH \rightarrow NaClO_{4} + H_2O\] Since the balanced chemical equation shows a 1:1 mole ratio between HClO4 and NaOH, we can use the given molarity of NaOH to find the moles of NaOH: moles of NaOH = Molarity of NaOH × volume of NaOH = 0.0875 M × 50.00 mL = 4.375 mmol Now, using the moles of NaOH, we can find the volume of HClO4 needed: volume of HClO4 = (moles of NaOH / Molarity of HClO4) = (4.375 mmol / 0.115 M) = 38.043 mL Therefore, 38.043 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH.

First, let's write the balanced chemical equation for the reaction between HCl and Mg(OH)2: \[2HCl + Mg(OH)_{2} \rightarrow MgCl_{2} + 2H_2O\] Now, let's convert the mass of Mg(OH)2 given (2.87 g) into moles, using its molar mass (58.32 g/mol for Mg and 34.02 g/mol for 2 OH): moles of Mg(OH)2 = (2.87 g) / (58.32 g/mol + 2 × 34.02 g/mol) = 0.02479 mol Using the mole ratio from the balanced chemical equation (2:1), we can find the moles of HCl needed: moles of HCl = 2 × moles of Mg(OH)2 = 2 × 0.02479 mol = 0.04958 mol Now, using the given molarity of HCl (0.128 M), we can find the volume of HCl needed: volume of HCl = (moles of HCl / Molarity of HCl) = (0.04958 mol / 0.128 M) = 387.34 mL Therefore, 387.34 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2.

We are given the information that 25.8 mL of AgNO3 solution is needed to precipitate all Cl- ions from a 785-mg sample of KCl, forming AgCl. First, let's find the moles of Cl- ions in the 785-mg sample of KCl using its molar mass (39.10 g/mol for K and 35.45 g/mol for Cl): moles of Cl- = (0.785 g) / (39.10 g/mol + 35.45 g/mol) = 0.01049 mol Now, we can write the balanced chemical equation for the reaction between AgNO3 and KCl: \[AgNO_{3} + KCl \rightarrow AgCl + KNO_{3}\] Since the balanced chemical equation shows a 1:1 mole ratio between AgNO3 and Cl-, the moles of AgNO3 required for the reaction is equal to the moles of Cl- ions. moles of AgNO3 = moles of Cl- = 0.01049 mol Now, using the given volume of the AgNO3 solution (25.8 mL), we can find the molarity of AgNO3: Molarity of AgNO3 = (moles of AgNO3 / volume of AgNO3) = (0.01049 mol / 25.8 mL) = 0.4065 M Therefore, the molarity of the AgNO3 solution is 0.4065 M.

We are given the information that 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH. First, let's find the moles of HCl using its molarity and volume: moles of HCl = Molarity of HCl × volume of HCl = 0.108 M × 45.3 mL = 4.8876 mmol Now, let's write the balanced chemical equation for the reaction between HCl and KOH: \[HCl + KOH \rightarrow KCl + H_2O\] Since the balanced chemical equation shows a 1:1 mole ratio between HCl and KOH, the moles of KOH required for the reaction are equal to the moles of HCl: moles of KOH = moles of HCl = 4.8876 mmol Finally, using the molar mass of KOH (56.11 g/mol), we can convert the moles of KOH to grams: mass of KOH = moles of KOH × molar mass of KOH = 4.8876 mmol × 56.11 g/mol = 0.2742 g Therefore, there are 0.2742 g of KOH present in the solution.