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Chapter 18: Problem 50

Will the following precipitates form under the given conditions? (a) \(\mathrm{PbI}_{2}(\mathrm{s}),\) from a solution that is \(1.05 \times 10^{-3} \mathrm{M} \mathrm{HI}\), \(1.05 \times 10^{-3} \mathrm{M} \mathrm{NaI},\) and \(1.1 \times 10^{-3} \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). (b) \(\operatorname{Mg}(\mathrm{OH})_{2}(\mathrm{s}),\) from \(2.50 \mathrm{L}\) of \(0.0150 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) to which is added 1 drop \((0.05 \mathrm{mL})\) of \(6.00 \mathrm{M} \mathrm{NH}_{3}\). (c) \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) from a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Al}^{3+}, 0.010 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH},\) and \(0.010 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\).

Short Answer

Expert verified
(a) Yes, a precipitate will form. (b) No, a precipitate will not form. (c) No, a precipitate will not form.

Step by step solution

01

Find Ksp values

Firstly, we need to find the solubility product constant (Ksp) values for the three compounds PbI2, Mg(OH)2, and Al(OH)3 from a solubility chart. These constants are given as follows: PbI2 = 7.1 x 10^-9, Mg(OH)2 = 5.61 x 10^-12, Al(OH)3 = 1.9 x 10^-33.
02

Calculate Ion Product (Q) for PbI2

Now calculate the Ion Product (Q) for the first solution (a). To do this, multiply the molar concentrations of the ions that make up PbI2. So, Q for PbI2 is (1.1 x 10^-3) x (1.05 x 10^-3)^2 = 1.22 x 10^-9
03

Compare Q and Ksp for PbI2

Compare the calculated Q value with the Ksp for PbI2. Since Q (1.22 x 10^-9) is greater than the Ksp (7.1 x 10^-9), a precipitate will form.
04

Calculate Q for Mg(OH)2

Like step 2, we calculate Q for Mg(OH)2 of solution (b). For this one, you must account for the volume of the entire solution as well as the contribution of NH3 to [OH-]. Q for Mg(OH)2 = (0.0150 M)^2 * [OH-]^2. As [OH-] is a component of NH3, we need to calculate its concentration in the solution. The amount of NH3 = 6 M * 0.05 mL = 0.3 mmol. As this is distributed in 2.50 L, [OH-] = 0.3 mmol / 2.50 L = 0.12 M. Plug this into the equation, we get Q = (0.015)^2 * (0.12)^2 = 3.24 x 10^-6.
05

Compare Q and Ksp for Mg(OH)2

Compare Q and Ksp for Mg(OH)2. Since Q (3.24 x 10^-6) is less than Ksp (5.61 x 10^-12), no precipitate forms.
06

Calculate Q for Al(OH)3

Repeat for solution (c) to calculate Q for Al(OH)3. Both CH3COOH and NaCH3COO are sources of OH-. Here [OH-]= 2 * 0.01 = 0.02. Using the formula for Al(OH)3, Q = (0.01)^3 * (0.02)^3 = 8 x 10^-12.
07

Compare Q and Ksp for Al(OH)3

Compare Q and Ksp for Al(OH)3. Since Q (8 x 10^-12) is less than Ksp (1.9 x 10^-33), no precipitate forms.

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Most popular questions from this chapter

Assume that, to be visible to the unaided eye, a precipitate must weigh more than \(1 \mathrm{mg}\). If you add \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})\) to \(100.0 \mathrm{mL}\) of a clear saturated aqueous AgCl solution, will you be able to see \(\mathrm{AgCl}(\mathrm{s})\) precipitated as a result of the common-ion effect? Explain.

Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

An aqueous solution that \(2.00 \mathrm{M}\) in \(\mathrm{AgNO}_{3}\) is slowly added from a buret to an aqueous solution that is \(0.0100 \mathrm{M}\) in \(\mathrm{Cl}^{-}\) and \(0.250 \mathrm{M}\) in \(\mathrm{I}^{-}\). (a) Which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is the first to precipitate? (b) When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c) Is the separation of \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\) feasible by fractional precipitation in this solution?

The best way to ensure complete precipitation from saturated \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) of a metal ion, \(\mathrm{M}^{2+}\), as its sulfide, \(\mathrm{MS}(\mathrm{s}),\) is to \((\mathrm{a})\) add an acid; \((\mathrm{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{S}\right]\) in the solution; (c) raise the \(\mathrm{pH} ;\) (d) heat the solution.

A solution is \(0.05 \mathrm{M}\) in \(\mathrm{Cu}^{2+},\) in \(\mathrm{Hg}^{2+},\) and in \(\mathrm{Mn}^{2+}\). Which sulfides will precipitate if the solution is made to be \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) and \(0.010 \mathrm{M} \mathrm{HCl}(\mathrm{aq}) ?\) For \(\mathrm{CuS}\), \(K_{\mathrm{spa}}=6 \times 10^{-16} ;\) for \(\mathrm{HgS}, K_{\mathrm{spa}}=2 \times 10^{-32} ;\) for \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).
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