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Chapter 4: Problem 72

Barium hydroxide, often used to titrate weak organic acids, is obtained as the octahydrate, \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\). What mass of \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) would be required to make \(500.0 \mathrm{~mL}\) of a solution that is \(0.1500 \mathrm{M}\) in hydroxide ions?

Short Answer

Expert verified
11.83 g of \(\mathrm{Ba(OH)_2 . 8H_2O}\) are required to make 500 mL of a solution with a molarity of 0.1500 M in hydroxide ions.

Step by step solution

01

Conversion of Volume

Convert the volume of the solution from milliliters to liters by dividing by 1000: \(500.0 \, \mathrm{mL} = 0.500 \, \mathrm{L}\) since there are 1000 milliliters in one liter.
02

Calculate the moles of Hydroxide ions required

Calculate the required moles of hydroxide ions: Molarity (\(M\)) = moles/solution volume in liters;Rearrange for moles = Molarity (\(M\)) * Solution Volume (\(L\)) = \(0.1500 \, \mathrm{M} \times 0.500 \, \mathrm{L} = 0.075 \, \mathrm{moles}\) of \(\mathrm{OH}^{-}\)
03

Calculate moles of \(\mathrm{Ba(OH)_2 . 8H_2O}\)

For every molecule of \(\mathrm{Ba(OH)_{2} . 8H_2O}\), there are two hydroxide ions. Hence divide the required moles of \(\mathrm{OH}^{-}\) by 2 to get moles of \(\mathrm{Ba(OH)_{2} . 8H_2O} = 0.075 \, \mathrm{moles} \, \mathrm{OH}^{-} / 2 = 0.0375 \, \mathrm{moles}\)
04

Calculate the mass of \(\mathrm{Ba(OH)_2 . 8H_2O}\)

Convert moles to mass using the molar mass of \(\mathrm{Ba(OH)_{2} . 8H_2O}\). The molar mass of barium is 137.33 g/mol, the molar mass of oxygen is 15.999 g/mol, and the molar mass of hydrogen is 1.007 g/mol. So the molar mass of \(\mathrm{Ba(OH)_{2} . 8H_2O = 137.33 + (2 \times (15.999 + 1.007)) + (8 \times (2 \times 1.007 +15.999)) = 315.47 \, \mathrm{g/mol}\). Mass = moles \(\times\) molar mass = \(0.0375 \, \mathrm{moles} \times 315.47 \, \mathrm{g/mol} = 11.83 \, \mathrm{g} \)

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Most popular questions from this chapter

Classify the following reactions according to the types discussed in the chapter. (a) \(\mathrm{Cl}_{2}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO}_{3}\) (c) \(\mathrm{NH}_{3}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{4}^{+}\) (d) \(2 \mathrm{CCl}_{4}+\mathrm{CrO}_{4}^{2-} \longrightarrow\) \(2 \mathrm{COCl}_{2}+\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{Cl}^{-}\) (e) \(\mathrm{Ca}+\mathrm{F}_{2} \longrightarrow \mathrm{CaF}_{2}\) (f) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (g) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{BaSO}_{4}\) (h) \(\mathrm{CuO}+\mathrm{H}_{2} \longrightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (i) \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (j) \(2 \mathrm{FeCl}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}\) (k) \(\mathrm{LiOH}+\mathrm{HNO}_{3} \longrightarrow \mathrm{LiNO}_{3}+\mathrm{H}_{2} \mathrm{O}\)

A 0.9157 -g mixture of \(\mathrm{CaBr}_{2}\) and \(\mathrm{NaBr}\) is dissolved in water, and \(\mathrm{AgNO}_{3}\) is added to the solution to form AgBr precipitate. If the mass of the precipitate is \(1.6930 \mathrm{~g}\), what is the percent by mass of \(\mathrm{NaBr}\) in the original mixture?

Based on oxidation number consideration, explain why carbon monoxide (CO) is flammable but carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is not.

Explain why potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\) and potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)\) can serve as internal indicators in redox titrations.

Describe the basic steps involved in an acid-base titration. Why is this technique of great practical value?
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