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Chapter 16: Problem 12

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

Short Answer

Expert verified
When \(Na_{3}PO_{4}(aq)\) is added to a solution containing a metal ion, the two possible precipitates that can form are: 1. Metal Phosphate (MP) with a general formula of \(M_{x}(PO_{4})_{y}\) 2. Metal Hydrogen Phosphate (MHP) with a general formula of \(M_{x}(HPO_{4})_{y}\)

Step by step solution

01

Identify the ions in the sodium phosphate solution

The aqueous sodium phosphate (Na3PO4) solution contains sodium ions (Na+) and phosphate ions (PO4^3-).
02

Determine the possible precipitate reactions

When a phosphate solution is combined with a solution containing a metal ion, the phosphate ions (PO4^3-) can react with the metal ions to form a precipitate. There are typically two possible types of precipitates that can form: metal phosphate (MP) and metal hydrogen phosphate (MHP). Type 1: Metal Phosphate (MP) A metal phosphate is formed when the phosphate ion (PO4^3-) directly reacts with the metal ion, with the general formula being Mx(PO4)y, where M is the metal ion and x and y depend on the charges of the metal ion and phosphate ion. Type 2: Metal Hydrogen Phosphate (MHP) A metal hydrogen phosphate is formed when the phosphate ions (PO4^3-) partially react with the metal ions to form Mx(HPO4)y, where M is the metal ion, and x and y depend on the charges of the metal ion and hydrogen phosphate ion (HPO4^2-).
03

Conclude with the two general possibilities for precipitate formations

When Na3PO4(aq) is added to a solution containing metal ions, the two possible types of precipitates that can form are: 1. Metal Phosphate (MP) with a general formula of Mx(PO4)y 2. Metal Hydrogen Phosphate (MHP) with a general formula of Mx(HPO4)y

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Most popular questions from this chapter

The stepwise formation constants for a complex ion usually have values much greater than \(1 .\) What is the significance of this?

The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: \(\mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}^{+}(a q) \quad K_{1}=1.0 \times 10^{\mathrm{s}}\) \(\mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{2}(a q) \quad K_{2}=1.0 \times 10^{5}\) \(\mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{3}-(a q) \quad K_{3}=1.0 \times 10^{\circ}\) \(\mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q) \quad K_{4}=1.0 \times 10^{\mathrm{s}}\) A solution is prepared by dissolving \(0.088\) mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(5.00\) mole of Nal in enough water to make \(1.0 \mathrm{~L}\) of solution. a, Calculate the equilibrium concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right]\). b. Calculate the equilibrium concentration of \(\left[\mathrm{I}^{-}\right]\). c. Calculate the equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right]\).

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}\). Calculate the \(K_{\text {sp }}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

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 copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times\) \(10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : \(\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}\) a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 M\)
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