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Chapter 16: Problem 47

Will a precipitate form when \(100.0 \mathrm{~mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{~mL}\) of \(2.0 \times 10^{-4} \mathrm{M} \mathrm{NaOH} ?\)

Short Answer

Expert verified
A precipitate will not form when \(100.0 \mathrm{~mL}\) of \(4.0 \times 10^{-4} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{~mL}\) of \(2.0 \times 10^{-4} M \mathrm{NaOH}\) as the calculated reaction quotient (Q) of \(2.0 \times 10^{-12}\) is less than the Ksp (\(1.5 \times 10^{-11}\)) for \(Mg(OH)_2\).

Step by step solution

01

Write the balanced equation

Write the balanced equation for the potential precipitation reaction: \[Mg^{2+} + 2OH^- \rightarrow Mg(OH)_2\]
02

Determine the initial concentrations of ions

Determine the initial concentrations of the ions before the solutions are mixed: \[Mg^{2+} = 4.0 \times 10^{-4} M\] \[OH^- = 2.0 \times 10^{-4} M\]
03

Calculate the final concentrations of ions

Calculate the final concentrations of the ions after mixing the two solutions. The volume has doubled, so the concentrations are halved: \[Mg^{2+} = \frac{4.0 \times 10^{-4} M}{2} = 2.0 \times 10^{-4} M\] \[OH^- = \frac{2.0 \times 10^{-4} M}{2} = 1.0 \times 10^{-4} M\]
04

Calculate the reaction quotient (Q)

Calculate the reaction quotient (Q) using the concentrations of Mg2+ and OH- ions: \[Q = [Mg^{2+}][OH^-]^2\] \[Q = (2.0 \times 10^{-4})(1.0 \times 10^{-4})^2\] \[Q = 2.0 \times 10^{-12}\]
05

Compare Q to Ksp

Compare the reaction quotient (Q) to the Ksp of Mg(OH)2. The Ksp for Mg(OH)2 is approximately \(1.5 \times 10^{-11} \). If Q > Ksp, a precipitate will form; if Q < Ksp, no precipitate will form: \[Q = 2.0 \times 10^{-12} < Ksp = 1.5 \times 10^{-11}\] Since Q is less than Ksp, no precipitate will form upon mixing the two solutions.

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Most popular questions from this chapter

The copper(I) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: \(\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K_{\mathrm{f}}=1.0 \times 10^{11}\) a. Calculate the solubility of \(\operatorname{CuBr}(s)\left(K_{\text {sp }}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\).b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\). EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00 .\)

A solution is prepared by mixing \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{~mL}\) of \(1.0 M \mathrm{KCl}\). Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\text {sp }}\right.\) for \(\mathrm{PbCl}_{2}(s)\) is \(\left.1.6 \times 10^{-5} .\right]\)
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