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Chapter 18: Problem 20

If \(100.0 \mathrm{mL}\) of \(0.0025 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is saturated with \(\mathrm{CaSO}_{4},\) how many grams of \(\mathrm{CaSO}_{4}\) would be present in the solution? [Hint: Does the usual simplifying assumption hold?]

Short Answer

Expert verified
The solution would contain 0.034 grams of \(\mathrm{CaSO}_{4}\).

Step by step solution

01

Calculate the molar mass of \(\mathrm{CaSO}_{4}\)

Add the atomic masses of calcium, sulfur, and four oxygen atoms to get the molar mass of \(\mathrm{CaSO}_{4}\): \(\mathrm{Molar \: mass\: of\: CaSO}_{4}=40.08+32.06+4*16.00=136.14 \: \mathrm{g/mol}\)
02

Determine the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) present in the solution

Use the molarity and volume of the solution to find the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\):\(\mathrm{Moles \: of\: Na}_{2}\mathrm{SO}_{4}=\mathrm{Molarity}\times\mathrm{Volume}=0.0025\: M\times 0.100\: L=0.00025\: \mathrm{moles}\)Given the stoichiometry of the reaction, the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) equal the moles of \(\mathrm{CaSO}_{4}\). So, you have 0.00025 moles of \(\mathrm{CaSO}_{4}\).
03

Convert moles of \(\mathrm{CaSO}_{4}\) to grams

Use the molar mass of \(\mathrm{CaSO}_{4}\) calculated in Step 1 to convert the moles of \(\mathrm{CaSO}_{4}\) to grams:\(\mathrm{Grams \: of\: CaSO}_{4}=\mathrm{Moles\: of\: CaSO}_{4}\times \mathrm{molar \: mass\: of\: CaSO}_{4}=0.00025\: \mathrm{moles}\times 136.14\: \mathrm{g/mol}=0.034\: \mathrm{g}\)

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Most popular questions from this chapter

A solution is \(0.10 \mathrm{M}\) in free \(\mathrm{NH}_{3}, 0.10 \mathrm{M}\) in \(\mathrm{NH}_{4} \mathrm{Cl}\), and \(0.015 \mathrm{M}\) in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} .\) Will \(\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate from this solution? \(K_{\mathrm{sp}}\) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) is \(2.2 \times 10^{-20}\).

Which one of the following solutions can be used to separate the cations in an aqueous solution in which \(\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.050 \mathrm{M}: 0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq}), 0.05 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\) \(0.001 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}),\) or \(0.50 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\text { aq }) ?\) Explain why.

Briefly describe each of the following ideas, methods, or phenomena: (a) common-ion effect in solubility equilibrium; (b) fractional precipitation; (c) ion-pair formation; (d) qualitative cation analysis.

Determine the molar solubility of lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2},\) in a buffer solution with \(\mathrm{pH}=3.00,\) given that \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=2.5 \times 10^{-9}\) \(\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{a}}=1.9 \times 10^{-5}\)

Saturated solutions of sodium phosphate, copper(II) chloride, and ammonium acetate are mixed together. The precipitate is (a) copper(II) acetate; (b) copper(II) phosphate; (c) sodium chloride; (d) ammonium phosphate; (e) nothing precipitates.
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