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Chapter 2: Problem 9

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. If the experiment were repeated and the \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) was diluted to 40.0 \(\mathrm{mL}\) with distilled water prior to the titration, how would that affect the volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) needed to reach the equivalence point? (A) It would be cut in half. (B) It would decrease by a factor of 1.5. (C) It would double. (D) It would not change.

Short Answer

Expert verified
The volume of Ca(NO3)2 needed to reach the equivalence point would not change. Hence, the answer is (D).

Step by step solution

01

Understand the initial reaction

In the initial titration, you have 20 mL of 1 M Na2CO3 solution. To calculate the amount of moles of sodium carbonate, use the formula volume (in liters) * molarity (in moles/LITER). Hence, moles of Na2CO3 are \(0.020 * 1 = 0.02\) moles.
02

Determine the Reaction Ratio

Sodium carbonate (Na2CO3) reacts with calcium nitrate (Ca(NO3)2) to form a precipitate. The reaction has the chemical equation \( Na2CO3 + Ca(NO3)2 \rightarrow 2NaNO3 + CaCO3 \). In this reaction, 1 mole of Na2CO3 reacts with 1 mole of Ca(NO3)2, hence we can infer that the initial titration required 0.02 moles of Ca(NO3)2.
03

Understand the Impact of Dilution

When the experiment is repeated with the Na2CO3 solution diluted to 40 mL, the molarity of the Na2CO3 solution is halved. However, the total number of moles of Na2CO3 remains the same (since dilution doesn't alter the quantity of solute). Therefore, the volume of 1 M Ca(NO3)2 required to reach the equivalence point will remain the same.

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Key Concepts

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

Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is the quantitative relationship between the amounts of substances involved. For example, by using stoichiometry, we can determine how much of a reactant is needed to completely react with another substance, or how much product will form.
In a balanced chemical equation, the coefficients indicate the ratio in which the substances react and form.
  • This ratio is critical for accurately calculating the required amounts of substances in a reaction.
  • For instance, in the reaction between Na\(_2\)CO\(_3\) and Ca(NO\(_3\))\(_2\), we see that 1 mole of Na\(_2\)CO\(_3\) reacts with 1 mole of Ca(NO\(_3\))\(_2\).
Understanding this helps us to predict the outcome of a reaction and manage reactant quantities effectively.
Molarity
Molarity (M) is a measurement of the concentration of a solution. It is defined as the number of moles of solute per liter of solution.
This concept is vital in titration experiments, where precise concentrations are needed to reach equivalence points accurately.
  • The formula to calculate molarity is: \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]
  • In the provided exercise, the molarity of Na\(_2\)CO\(_3\) before dilution was 1.0 M, which means there was 1 mole of Na\(_2\)CO\(_3\) in every liter of solution.
Knowing the molarity helps in determining how much of another solution is needed to reach the equivalence point.
Equivalence Point
The equivalence point in a titration is when the amount of titrant is just enough to completely react with the analyte or substance being tested. It's a crucial moment in a titration because it indicates when the reactants have reacted in perfect stoichiometric proportion.
In our exercise, the equivalence point is achieved when there is enough Ca(NO\(_3\))\(_2\) to completely react with the Na\(_2\)CO\(_3\).
  • At this point, the stoichiometric amount of each reactant has been utilized.
  • When dealing with a 1:1 mole ratio, as is the case here, determining the equivalence point becomes straightforward.
Reaching the equivalence point means that all the sodium carbonate originally present has reacted to form a precipitate.
Dilution
Dilution is the process of reducing the concentration of a solute in a solution, often by adding more solvent.
Even though dilution changes the concentration, it does not alter the total number of moles of solute in the solution.
  • This means that dilution will affect the molarity of a solution but not the total quantity of the solute present.
  • In the exercise, diluting the Na\(_2\)CO\(_3\) solution from 20 mL to 40 mL halved its molarity but maintained the total moles of Na\(_2\)CO\(_3\).
This concept explains why the volume of Ca(NO\(_3\))\(_2\) required to reach the equivalence point remains unchanged despite the dilution of Na\(_2\)CO\(_3\).

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Most popular questions from this chapter

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. How much \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to reach the equivalence point? (A) 10.0 mL (B) 20.0 mL (C) 30.0 mL (D) 40.0 mL

Hydrogen fluoride, HF, is a liquid at \(15^{\circ} \mathrm{C}\) . All other hydrogen halides (represented by HX, where \(\mathrm{X}\) is any other halogen) are gases at the same temperature. Why? (A) Fluorine has a very high electronegativity; therefore, the H鈥揊 bond is stronger than any other H鈥揦 bond. (B) HF is smaller than any other H鈥揦 molecule; therefore, it exhibits stronger London dispersion forces. (C) The dipoles in a HF molecule exhibit a particularly strong attraction force to the dipoles in other HF molecules. (D) The H鈥揊 bond is the most ionic in character compared to all other hydrogen halides.

Consider the Lewis structures for the following molecules: $$\begin{equation} \mathrm{CO}_{2}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-}, \text {and } \mathrm{NO}_{3}^{-} \end{equation}$$ Which molecule would have the shortest bonds? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{CO}_{3}^{2-}\) (C) \(\mathrm{NO}_{2}^{-}\) (D) \(\mathrm{NO}_{3}^{-}\)

Consider the Lewis structures for the following molecules: $$\begin{equation} \mathrm{CO}_{2}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-}, \text {and } \mathrm{NO}_{3}^{-} \end{equation}$$ Which molecule or molecules exhibit \(s p^{2}\) hybridization around the central atom? (A) \(\mathrm{CO}_{2}\) and \(\mathrm{CO}_{3}^{2-}\) (B) \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3^{-}}\) (C) \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{NO}_{3}^{-}\) (D) \(\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-}\)

\(2 \mathrm{Ag}^{+}(a q)+\mathrm{Fe}(s) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Fe}^{2+}(a q)\) Which of the following would cause an increase in potential in the voltaic cell described by the above reaction? (A) Increasing \(\left[\mathrm{Fe}^{2+}\right]\) (B) Adding more \(\mathrm{Fe}(s)\) (C) Decreasing \(\left[\mathrm{Fe}^{2+}\right]\) (D) Removing some \(\mathrm{Fe}(s)\)
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