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Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reaction. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{AgNO}_{3}\), (b) \(\mathrm{NaNO}_{3}\) and \(\mathrm{NiSO}_{4}\), (c) \(\mathrm{FeSO}_{4}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

Short Answer

Expert verified
(a) Precipitation will occur with the formation of Ag2CO3 (silver carbonate) solid. The balanced chemical equation is: \(Na_{2}CO_{3}(aq) + 2AgNO_{3}(aq) \rightarrow 2NaNO_{3}(aq) + Ag_{2}CO_{3}(s)\) (b) No precipitation will occur when NaNO3 and NiSO4 solutions are mixed. (c) Precipitation will occur with the formation of PbSO4 (lead sulfate) solid. The balanced chemical equation is: \(FeSO_{4}(aq) + Pb(NO_{3})_{2}(aq) \rightarrow Fe(NO_{3})_{2}(aq) + PbSO_{4}(s)\)

Step by step solution

01

(a) Mixing Na2CO3 and AgNO3 solutions

To determine if precipitation occurs when Na2CO3 (sodium carbonate) and AgNO3 (silver nitrate) solutions are mixed, we should first consider the double displacement reaction between the two compounds: Na2CO3(aq) + 2AgNO3(aq) -> 2NaNO3(aq) + Ag2CO3(s) Now we need to check the solubility of the products using solubility rules. Sodium nitrate (NaNO3) is soluble in water because nitrates are generally soluble. However, silver carbonate (Ag2CO3) is insoluble, as most carbonates are insoluble. Hence, precipitation will occur since Ag2CO3 will form a solid precipitate. The balanced chemical equation for the reaction is: Na2CO3(aq) + 2AgNO3(aq) -> 2NaNO3(aq) + Ag2CO3(s)
02

(b) Mixing NaNO3 and NiSO4 solutions

To determine if precipitation occurs when NaNO3 (sodium nitrate) and NiSO4 (nickel(II) sulfate) solutions are mixed, let's perform the double displacement reaction between the two compounds: NaNO3(aq) + NiSO4(aq) -> NaSO4(aq) + Ni(NO3)2(aq) Now we need to check the solubility of the products. Sodium sulfate (Na2SO4) is soluble in water since most sulfates are soluble. Nickel(II) nitrate [Ni(NO3)2] is also soluble because nitrates are generally soluble. Since both products are soluble in water, no precipitation will occur, and no further balanced equation is needed.
03

(c) Mixing FeSO4 and Pb(NO3)2 solutions

To determine if precipitation occurs when FeSO4 (iron(II) sulfate) and Pb(NO3)2 (lead(II) nitrate) solutions are mixed, let's perform the double displacement reaction between the two compounds: FeSO4(aq) + Pb(NO3)2(aq) -> Fe(NO3)2(aq) + PbSO4(s) Now we need to check the solubility of the products. Iron(II) nitrate [Fe(NO3)2] is soluble in water since nitrates are generally soluble. However, lead sulfate (PbSO4) is insoluble, as most sulfates of heavy metals (like lead) are insoluble. Hence, precipitation will occur since PbSO4 will form a solid precipitate. The balanced chemical equation for the reaction is: FeSO4(aq) + Pb(NO3)2(aq) -> Fe(NO3)2(aq) + PbSO4(s)

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Key Concepts

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

Double Displacement Reaction
In the realm of chemistry, one of the fundamental reaction types is the double displacement reaction, also known as a metathesis reaction. It occurs when components from two compounds exchange places to form two new compounds. This typically happens when two ionic aqueous solutions mix, and the anions and cations interchange partners.

For instance, when solutions of Na2CO3 (sodium carbonate) and AgNO3 (silver nitrate) are combined, we witness a classic double displacement reaction. The sodium (Na+) ions pair up with nitrate (NO3-) ions, while the silver (Ag+) ions join forces with carbonate (CO32-) ions, resulting in NaNO3 (sodium nitrate) and Ag2CO3 (silver carbonate) as the products.

A key feature of these reactions is the potential formation of an insoluble compound, or precipitate, which can easily be identified as the solid that emerges from the liquid mixture. The ability to predict the formation of a precipitate is essential in the study of chemical reactions and has significant implications in various fields, including environmental science and pharmaceuticals.
Solubility Rules
Chemists have established solubility rules as a set of guidelines that predict the solubility of ionic compounds in water. Knowing these rules can help you determine whether a compound will dissolve to form an aqueous solution or remain undissolved as a solid precipitate when mixed with water.

Some general rules include:
  • Nitrates (NO3-) are almost always soluble.
  • Most sulfates (SO42-) are soluble, except those of heavy metals like lead (Pb2+).
  • Most carbonates (CO32-) are insoluble, with exceptions such as those of sodium (Na+) and potassium (K+).

Applying these rules to our earlier examples, we determine the solubility of the resulting compounds. In the case of Na2CO3 and AgNO3, we predicted the formation of Ag2CO3 precipitate because, according to the rules, carbonates are insoluble except for those of Na++. Similarly, in the reaction between FeSO4 and Pb(NO3)2, lead sulfate (PbSO4) was anticipated to be insoluble and therefore precipitate because lead is a heavy metal.
Balanced Chemical Equation
The principle of mass conservation is fundamental to the science of chemistry, dictating that in a chemical reaction, the mass of the reactants must equal the mass of the products. To satisfy this principle, we represent reactions using balanced chemical equations. In such equations, the same number of atoms for each element is present on both sides of the reaction.

Let's break down the balancing process:
  • Write down the unbalanced equation with the correct formulas for all reactants and products.
  • Count the number of atoms for each element present in the reactants and products.
  • Add coefficients in front of the chemical formulas to balance the number of atoms for each element.
  • Check that the total charge is the same on both sides of the equation if the reaction involves ionic compounds.

Returning to our previous examples, the chemical equation Na2CO3(aq) + 2AgNO3(aq) -> 2NaNO3(aq) + Ag2CO3(s) is balanced because there are equal numbers of Na, Ag, N, C, and O atoms on both sides of the equation. It also reflects the physical state of each compound, with (aq) denoting aqueous solutions and (s) signifying solid precipitates. This complete representation of the reaction is essential for understanding the process's stoichiometry and practical applications, such as calculating the yield in a chemical synthesis.

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Most popular questions from this chapter

Indicate the concentration of each ion present in the solution formed by mixing (a) \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) and \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\), (b) \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\), (c) \(3.60 \mathrm{~g} \mathrm{KCl}\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution, \((\mathrm{c}) 0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

Using solubility guidelines, predict whether each of the following compounds is soluble or insoluble in water: (a) \(\mathrm{NiCl}_{2}\) (b) \(\mathrm{Ag}_{2} \mathrm{~S},(\mathrm{c}) \mathrm{Cs}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{SrCO}_{3}\) (e) \(\mathrm{PbSO}_{4}\).

We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? [Section 4.2]

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