(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of $0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}$ solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If $55.8 \mathrm{~mL}\( of a \)\mathrm{BaCl}_{2}$ solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of $\mathrm{Na}_{2} \mathrm{SO}_{4}\(, what is the molarity of the \)\mathrm{BaCl}_{2}$ solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{MHCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

Step by step solution

01

Balance the chemical equation

The balanced chemical equation for the neutralization reaction between HCl and Ba(OH)2 is: \[2 \mathrm{HCl} + \mathrm{Ba(OH)_2} \rightarrow \mathrm{BaCl_2} + 2 \mathrm{H_2O}\]

02

Calculate the moles of Ba(OH)2

Using the given volume and concentration of Ba(OH)2, we can find the moles of Ba(OH)2: Moles of Ba(OH)2 = volume × molarity Moles of Ba(OH)2 = 50.0 mL × 0.101 M = 0.00505 mol

03

Find the moles of HCl needed

From the balanced equation, we know that 2 moles of HCl react with 1 mole of Ba(OH)2. Therefore, moles of HCl needed = 2 × moles of Ba(OH)2 = 2 × 0.00505 mol = 0.0101 mol

04

Calculate the volume of HCl required

Using the moles of HCl and its molarity, we can find the volume of HCl needed: Volume of HCl = moles of HCl / molarity Volume of HCl = 0.0101 mol / 0.120 M = \(83.75 \mathrm{~mL}\) (b)

05

Balance the chemical equation

The balanced chemical equation for the neutralization reaction between H2SO4 and NaOH is: \[2 \mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2 \mathrm{H_2O}\]

06

Calculate the moles of NaOH

Using the given mass and molar mass of NaOH (40.00 g/mol), we can find the moles of NaOH: Moles of NaOH = mass / molar mass Moles of NaOH = 0.200 g / 40.00 g/mol = 0.00500 mol

07

Find the moles of H2SO4 needed

From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, moles of H2SO4 needed = moles of NaOH / 2 = 0.00500 mol / 2 = 0.00250 mol

08

Calculate the volume of H2SO4 required

Using the moles of H2SO4 and its molarity, we can find the volume of H2SO4 needed: Volume of H2SO4 = moles of H2SO4 / molarity Volume of H2SO4 = 0.00250 mol / 0.125 M = \(20.0 \mathrm{~mL}\) (c)

09

Balance the chemical equation

The balanced chemical equation for the precipitation reaction between Na2SO4 and BaCl2 is: \[\mathrm{Na}_2\mathrm{SO}_4 + \mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 \downarrow+ 2 \mathrm{NaCl}\]

10

Calculate the moles of Na2SO4

Using the given mass and molar mass of Na2SO4 (142.04 g/mol), we can find the moles of Na2SO4: Moles of Na2SO4 = mass / molar mass Moles of Na2SO4 = 752 mg / 142.04 g/mol = 0.00530 mol (note: convert from mg to g)

11

Find the moles of BaCl2 needed

From the balanced equation, we know that 1 mole of BaCl2 reacts with 1 mole of Na2SO4. Therefore, moles of BaCl2 needed = moles of Na2SO4 = 0.00530 mol

12

Calculate the molarity of BaCl2

Using the moles of BaCl2 and the given volume, we can find the molarity of BaCl2: Molarity of BaCl2 = moles of BaCl2 / volume Molarity of BaCl2 = 0.00530 mol / 55.8 mL = \(0.0950 \mathrm{M}\) (note: convert from mL to L) (d)

13

Balance the chemical equation

The balanced chemical equation for the neutralization reaction between HCl and Ca(OH)2 is: \[2 \mathrm{HCl} + \mathrm{Ca(OH)_2} \rightarrow \mathrm{CaCl_2} + 2 \mathrm{H_2O}\]

14

Calculate the moles of HCl

Using the given volume and concentration of HCl, we can find the moles of HCl: Moles of HCl = volume × molarity Moles of HCl = 42.7 mL × 0.208 M = 0.00888 mol

15

Find the moles of Ca(OH)2 needed

From the balanced equation, we know that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, moles of Ca(OH)2 needed = moles of HCl / 2 = 0.00888 mol / 2 = 0.00444 mol

16

Calculate the mass of Ca(OH)2

Using the moles of Ca(OH)2 and its molar mass (74.09 g/mol), we can find the mass of Ca(OH)2: Mass of Ca(OH)2 = moles of Ca(OH)2 × molar mass Mass of Ca(OH)2 = 0.00444 mol × 74.09 g/mol = \(0.328 \mathrm{~g}\)

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