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Chapter 15: Problem 50

A solution contains \(1.0 \times 10^{-5} M \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} M \mathrm{CN}^{-}\) Will AgCN( \(s\) ) precipitate? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgCN}(s) \text { is } 2.2 \times 10^{-12} .\right)\)

Short Answer

Expert verified
Since the calculated reaction quotient, Q, (\(2.0 \times 10^{-11}\)) is greater than the solubility product constant, Ksp, (\(2.2 \times 10^{-12}\)), AgCN will precipitate in the given solution.

Step by step solution

01

Write out the reaction and Ksp expression

First, write out the reaction for the dissociation of AgCN(s) and the Ksp expression. Reaction: \(AgCN(s) \rightleftharpoons Ag^+(aq) + CN^-(aq)\) Ksp expression: \(K_{sp} = [Ag^+][CN^-]\) The given \(K_{sp}\) value for AgCN is \(2.2 \times 10^{-12}\).
02

Calculate the reaction quotient (Q)

Next, we can calculate the Q value using the given concentrations of Ag+ and CN- ions in the solution. Q = \([Ag^+]\)[\(CN^-\)\] Q = \( (1.0 \times 10^{-5})(2.0 \times 10^{-6})\) Now, calculate Q: Q = \( 2.0 \times 10^{-11}\)
03

Compare Q and Ksp

Now, we will compare the calculated Q value to the given Ksp value to determine if AgCN will precipitate in the solution. Ksp = \(2.2 \times 10^{-12}\) Q = \(2.0 \times 10^{-11}\) Since Q > Ksp (\(2.0 \times 10^{-11}\) > \(2.2 \times 10^{-12}\)), we can conclude that AgCN will precipitate in the given solution.

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Most popular questions from this chapter

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10 M \operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) NaI

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{CaF}_{2}(s), K_{\mathrm{sp}}=4.0 \times 10^{-11},\) or \(\mathrm{BaF}_{2}(s), K_{\mathrm{sp}}=2.4 \times 10^{-5}\) b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s), K_{\mathrm{sp}}=1.3 \times 10^{-32},\) or \(\mathrm{FePO}_{4}(s)\) \(K_{\mathrm{sp}}=1.0 \times 10^{-22}\)

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\).

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{Hg} \mathrm{I}_{4}^{2-} .\) )

Calculate the solubility of \(\operatorname{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a pH of \(11.00 .\)
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