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Ignoring protolysis reactions (i.e. proton transfer reaction), specify what ions are present in a solution upon dissolving each of the following substances in water: (a) \(\mathrm{Li}_{2} \mathrm{CO}_{3},\) (b) \((\mathrm{NH} 4)_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{NaPF}_{6^{\circ}}\) (c) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Short Answer

Expert verified
In summary, the ions present in the solutions upon dissolving the given substances in water are: (a) \(Li_2CO_3\): Li+ and CO3^2- ions (b) \((NH_4)_3PO_4\): NH4+ and PO4^3- ions (c) \(NaPF_6\): Na+ and PF6^- ions (d) \(Na_2Cr_2O_7\): Na+ and Cr2O7^2- ions

Step by step solution

01

a) Li2CO3

In order to determine the ions produced when lithium carbonate (Li2CO3) is dissolved in water, we first look at the compound's formula. From this, we can extract the following ions: - Lithium ion (Li+) appears twice in the formula. Therefore, 2 Li+ ions will be produced per one unit of Li2CO3 dissolved. - Carbonate ion (CO3^2-) appears once in the compound. Upon dissolving Li2CO3 in water, the solution will contain Li+ ions and CO3^2- ions.
02

b) (NH4)3PO4

To find the ions formed when ammonium phosphate ((NH4)3PO4) dissolves in water, we will observe the compound formula. Based on this, we get: - Ammonium ion (NH4+) appears thrice. As a result, every (NH4)3PO4 molecule dissolved will create 3 NH4+ ions. - Phosphate ion (PO4^3-) is present once in the compound. When (NH4)3PO4 is dissolved in water, NH4+ ions and PO4^3- ions will be generated.
03

c) NaPF6

In order to discern the ions comprising the sodium hexafluorophosphate (NaPF6) when dissolved in water, we need to dissect its formula. Per the formula, we have: - Sodium ion (Na+) is present once in the compound. - Hexafluorophosphate ion (PF6^-) is also present once in the compound. So, when NaPF6 is dissolved in water, the solution will have Na+ ions and PF6^- ions.
04

d) Na2Cr2O7

Lastly, we need to identify the ions present when sodium dichromate (Na2Cr2O7) dissolves. By examining its formula, we can identify the following ions: - Sodium ion (Na+) appears twice in the compound. Consequently, 2 Na+ ions will be formed per one molecule of Na2Cr2O7. - Dichromate ion (Cr2O7^2-) is present once in the compound. Upon dissolving Na2Cr2O7 in water, the solution will contain Na+ ions and Cr2O7^2- ions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion Formation
When ionic compounds dissolve in water, they separate into individual ions. This process is vital for many chemical reactions and is known as dissociation. Each ion is either positively charged (cation) or negatively charged (anion).
For instance:
  • Lithium carbonate (\(\mathrm{Li}_2 \mathrm{CO}_3\)) dissociates into two lithium ions (\(\mathrm{Li}^+\)) and one carbonate ion (\(\mathrm{CO}_3^{2-}\)).
  • Ammonium phosphate (\((\mathrm{NH}_4)_3\mathrm{PO}_4\)) breaks into three ammonium ions (\(\mathrm{NH}_4^+\)) and one phosphate ion (\(\mathrm{PO}_4^{3-}\)).
  • Sodium hexafluorophosphate (\(\mathrm{NaPF}_6\)) results in a sodium ion (\(\mathrm{Na}^+\)) and a hexafluorophosphate ion (\(\mathrm{PF}_6^-\)).
  • Sodium dichromate (\(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7\)) dissolves to give two sodium ions (\(\mathrm{Na}^+\)) and one dichromate ion (\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)).
Thus, ion formation is crucial in understanding the behavior of ionic compounds in aqueous solutions, allowing the ions to interact or participate in further chemical processes.
Chemical Formulas
Chemical formulas are like maps showing what an ionic compound is made of. They give us the necessary clues to predict the ions that will form when a compound dissolves.
  • An example is \(\mathrm{Li}_2 \mathrm{CO}_3\), where \(\mathrm{Li}_2\) indicates two lithium ions, and \(\mathrm{CO}_3\) shows the presence of the carbonate ion.
  • Similarly, \((\mathrm{NH}_4)_3\mathrm{PO}_4\) clearly shows three ammonium ions (\(\mathrm{NH}_4^+\)) will be produced for each molecule of the compound.
  • For \(\mathrm{NaPF}_6\), it's clear that the sodium ion (\(\mathrm{Na}^+\)) is paired with a hexafluorophosphate ion (\(\mathrm{PF}_6^-\)).
  • In \(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7\), the formula informs us about the two sodium ions and one dichromate ion.
Understanding chemical formulas is key to unlocking the mysteries of the ions that emerge upon dissolution. It is like knowing the recipe before you start cooking, ensuring you have all the right ingredients for the desired outcome.
Solubility in Water
Solubility refers to the ability of a substance to dissolve in a solvent, such as water. Not all ionic compounds are equally soluble in water. Whether a compound dissolves depends on the interactions between the water molecules and the ions in the compound.
When ionic compounds like \(\mathrm{Li}_2 \mathrm{CO}_3\), \((\mathrm{NH}_4)_3\mathrm{PO}_4\), \(\mathrm{NaPF}_6\), and \(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7\) dissolve in water, their ions become surrounded by water molecules. This process is known as hydration and it plays a crucial role in determining solubility.
  • The polar nature of water allows it to stabilize the ions, typically enhancing solubility.
  • Ionic compounds with ions that have similar sizes to water tend to be more soluble.
  • Solubility can vary greatly, and in some cases, external conditions like temperature can significantly affect it.
Solubility is thus essential to consider when predicting how and to what degree an ionic compound will dissolve, affecting its potential to participate in reactions or be used in solutions.

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Most popular questions from this chapter

As \(\mathrm{K}_{2} \mathrm{O}\) dissolves in water, the oxide ion reacts with water molecules to form hydroxide ions. (a) Write the molecular and net ionic equations for this reaction. (b) Based on the definitions of acid and base, what ion is the base in this reaction? (c) What is the acid in the reaction? (d) What is the spectator ion in the reaction?

The arsenic in a \(1.22-g\) sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in \(\mathrm{AsO}_{4}{ }^{3-} ?(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{Ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$

A \(1.248-g\) sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

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