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Chapter 18: Problem 95

The slightly soluble solute \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is most soluble in (a) pure water; (b) \(0.10 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4} ;\) (c) \(0.25 \mathrm{M} \mathrm{KNO}_{3}\); (d) \(0.40 \mathrm{M} \mathrm{AgNO}_{3}\).

Short Answer

Expert verified
(a) \(Ag_{2}CrO_{4}\) is most soluble in pure water, as there are no common ions present to reduce its solubility due to the common ion effect.

Step by step solution

01

Understanding Solubility and Common Ion Effect

Solubility is the amount of solute that can be dissolved in a solvent at a given temperature. The common ion effect refers to the decrease in solubility when a compound is mixed with a solution already containing one of its constituent ions. When \(Ag_{2}CrO_{4}\) dissolves, it dissociates into its ions \(2Ag^+\) and \(CrO_{4}^{2-}\).
02

Determining Solubility in Pure Water

In pure water, there are no common ions present. Consequently, there would be no restrictions on the solubility of the \(Ag_{2}CrO_{4}\) due to the common ion effect. So it can be dissolved to its maximum capacity.
03

Determining Solubility in \(0.10M K_{2}CrO_{4}\)

The \(K_{2}CrO_{4}\) solution contains the common ion \(CrO_{4}^{2-}\). Presence of this ion will reduce the solubility of \(Ag_{2}CrO_{4}\) in this solution due to the common ion effect.
04

Determining Solubility in \(0.25M KNO_{3}\)

The \(KNO_{3}\) solution does not have a common ion with \(Ag_{2}CrO_{4}\). However, it will not increase the solubility of \(Ag_{2}CrO_{4}\) as in the case of pure water. The solubility will be similar to that in pure water.
05

Determining Solubility in \(0.40M AgNO_{3}\)

In this case, \(AgNO_{3}\) shares a common ion \(Ag^+\) with \(Ag_{2}CrO_{4}\). Therefore, due to the common ion effect, the solubility of \(Ag_{2}CrO_{4}\) will be further reduced in this solution.

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Most popular questions from this chapter

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 \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).

Will AgI(s) precipitate from a solution with \(\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]=0.012 \mathrm{M}, \left[\mathrm{CN}^{-}\right]=1.05 \mathrm{M}, \) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\) For \( \mathrm{AgI}, K_{\mathrm{sp}}=8.5 \times 10^{-17} ; =\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, K_{\mathrm{f}}=5.6 \times 10^{18}\).

\(\mathrm{Cu}^{2+}\) and \(\mathrm{Pb}^{2+}\) are both present in an aqueous solution. To precipitate one of the ions and leave the other in solution, add (a) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) ;\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\); (c) \(\mathrm{HNO}_{3}(\mathrm{aq}) ;\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\).

The solubility of \(\mathrm{CdCO}_{3}(\mathrm{s})\) in \(1.00 \mathrm{M} \mathrm{KI}(\mathrm{aq})\) is \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Given that \(K_{\mathrm{sp}}\) of \(\mathrm{CdCO}_{3}\) is \(5.2 \times 10^{-12},\) what is \(K_{\mathrm{f}}\) for \(\left[\mathrm{CdI}_{4}\right]^{2-} ?\)

Lead(II) chloride is most soluble in (a) \(0.100 \mathrm{M} \mathrm{NaCl}\); (b) \(0.100 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ;(\mathrm{c}) 0.100 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ;(\mathrm{d}) 0.100 \mathrm{M}\) \(\mathrm{NaNO}_{3} ;(\mathrm{e}) 0.100 \mathrm{MnSO}_{4}\).
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