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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 \(\mathrm{Na}_3 \mathrm{PO}_4(aq)\) is added to a solution containing a metal ion, the two possible precipitates that can form are calcium phosphate \(\mathrm{Ca_3(PO_4)_2(s)}\) and aluminum phosphate \(\mathrm{AlPO_4(s)}\). This is because phosphate ions have low solubility with many metal ions, particularly calcium ions \(\mathrm{Ca}^{2+}\) and aluminum ions \(\mathrm{Al}^{3+}\).

Step by step solution

01

Identify the ions present in the sodium phosphate solution

In a sodium phosphate solution, we have sodium ions \(\mathrm{Na}^+\) and phosphate ions \(\mathrm{PO}_4^{3-}\). When the solution comes into contact with the solution containing the metal ions, the phosphate ions will most likely bond with the metal ions.
02

Understanding bonding with phosphate ions

Phosphate ions have a charge of -3, which means they can form ionic bonds with metal ions having positive charges. Metal ions usually have a charge of +1, +2, or +3.
03

Identifying common metal ions that form precipitates with phosphate ions

Based on solubility rules and common metal ions found in chemistry, the most common metal ions that form precipitates with phosphate ions are calcium ions \(\mathrm{Ca}^{2+}\) and aluminum ions \(\mathrm{Al}^{3+}\). These ions are known to form insoluble salts with phosphate ions.
04

Writing the chemical formulas for the precipitates

When the metal ions \(\mathrm{Ca}^{2+}\) and \(\mathrm{Al}^{3+}\) form precipitates with phosphate ions, the following precipitates are formed: 1. Calcium phosphate: \(\mathrm{Ca_3(PO_4)_2(s)}\) 2. Aluminum phosphate: \(\mathrm{AlPO_4(s)}\) In conclusion, the two possibilites for the precipitates when \(\mathrm{Na}_{3} \mathrm{PO}_{4}(aq)\) is added to a solution containing a metal ion are calcium phosphate \(\mathrm{Ca_3(PO_4)_2(s)}\) and aluminum phosphate \(\mathrm{AlPO_4(s)}\).

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Most popular questions from this chapter

You are browsing through the Handbook of Hypothetical Chemistry when you come across a solid that is reported to have a \(K_{\text {sp }}\) value of zero in water at \(25^{\circ} \mathrm{C}\). What does this mean?

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 \(2.0 \times 10^{-4} \mathrm{M} \mathrm{NaOH} ?\)

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}+\mathrm{P}_{3} \mathrm{O}_{10}^{5-} \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-} .\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L}\) of solution) after 40. g \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{~L}\) of the solution.

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\).

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 \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)
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