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Chapter 18: Problem 107

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 \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, K_{\mathrm{f}}=5.6 \times 10^{18}\).

Short Answer

Expert verified
The precipitation of AgI from the solution will depend on the comparison of the calculated Q value with the given Ksp value. If Q > Ksp, then AgI will precipitate.

Step by step solution

01

Calculate the concentration of Ag+ ions

The expression for the formation constant (Kf) for the complex ion is given by:\nKf = [\(\mathrm{Ag+}\)]/[\(\mathrm{Ag(CN)_{2}^{-}}\)][\(\mathrm{CN-}\)]².\nWe are given that Kf = \(5.6 \times 10^{18}\), [\(\mathrm{Ag(CN)_{2}^{-}}\)] = 0.012M and [\(\mathrm{CN-}\)] = 1.05M. Solving the equation for [\(\mathrm{Ag+}\)], we obtain:\n[\(\mathrm{Ag+}\)] = Kf × [\(\mathrm{Ag(CN)_{2}^{-}}\)] × [\(\mathrm{CN-}\)]²\nSubstituting the given values into this equation, we can calculate [\(\mathrm{Ag+}\)]
02

Compute the ionic product

To decide if AgI will precipitate, we have to calculate the ionic product for the AgI and compare it with the given solubility product (Ksp). The ionic product (Q) is given by Q = [\(\mathrm{Ag+}\)] × [\(\mathrm{I-}\)]. Using the calculated [\(\mathrm{Ag+}\)] from Step 1 and the given [\(\mathrm{I-}\)] = 2.0M, we can compute Q.
03

Compare ionic product with solubility product

The final step is to compare Q with Ksp. If Q > Ksp, then AgI will precipitate. So, we compare the calculated Q from Step 2 with the given Ksp = \(8.5 \times 10^{-17}\)

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Most popular questions from this chapter

Fluoridated drinking water contains about 1 part per million (ppm) of \(\mathrm{F}^{-}\). Is \(\mathrm{CaF}_{2}\) sufficiently soluble in water to be used as the source of fluoride ion for the fluoridation of drinking water? Explain. [Hint: Think of 1 ppm as signifying \(1 \mathrm{g} \mathrm{F}^{-}\) per \(10^{6} \mathrm{g}\) solution.

A handbook lists the solubility of \(\mathrm{CaHPO}_{4}\) as \(0.32 \mathrm{g}\) \(\mathrm{CaHPO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) and lists \(K_{\mathrm{sp}}\) as \(1 \times 10^{-7}\). (a) Are these data consistent? (That is, are the molar solubilities the same when derived in two different ways?) (b) If there is a discrepancy, how do you account for it?

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(1.00 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\).

Can \(\mathrm{Fe}^{2+}\) and \(\mathrm{Mn}^{2+}\) be separated by precipitating \(\mathrm{FeS}(\mathrm{s})\) and not \(\mathrm{MnS}(\mathrm{s}) ?\) Assume \(\left[\mathrm{Fe}^{2+}\right]=\left[\mathrm{Mn}^{2+}\right]=\) \(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.10 \mathrm{M} .\) Choose a \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) that ensures maximum precipitation of \(\mathrm{FeS}(\mathrm{s})\) but not \(\mathrm{MnS}(\mathrm{s}) .\) Will the separation be complete? For \(\mathrm{FeS}, K_{\mathrm{spa}}=6 \times 10^{2} ;\) for \(\mathrm{MnS}, K_{\mathrm{spa}}=3 \times 10^{7}\).

Which of the following solids is (are) more soluble in an acidic solution than in pure water: \(\mathrm{KCl}\), \(\mathrm{MgCO}_{3}\), \(\mathrm{FeS}, \mathrm{Ca}(\mathrm{OH})_{2,}\) or \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} ?\) Explain.
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