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Chapter 2: Problem 6

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

Short Answer

Expert verified
So, to reach the equivalence point, 20.0 mL of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added. Thus, the correct answer is (B) 20.0 mL.

Step by step solution

01

Write down the balanced chemical equation.

The reaction between sodium carbonate and calcium nitrate can be represented as: \\( \mathrm{Na}_{2}\mathrm{CO}_{3} + \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow 2\mathrm{NaNO}_{3} + \mathrm{CaCO}_{3} \\). This shows that one mole of sodium carbonate reacts with one mole of calcium nitrate to produce two moles of sodium nitrate and one mole of calcium carbonate precipitate.
02

Determine the amount of moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\).

Use the given volume and molar concentration of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) to calculate the number of moles. Molarity = Moles / Volume. Therefore, Moles = Molarity * Volume = 1.0 M * 0.020 L = 0.020 moles.
03

Use stoichiometry to find the needed \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\).

From the balanced chemical equation, the stoichiometric ratio (or mole ratio) between \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) and \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is 1:1. Thus, to neutralize 0.020 moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\), you need exactly the same amount, 0.020 moles, of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\).
04

Calculate the volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) required.

To find the needed volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\), use the molarity formula again but rearranged to solve for volume. Volume = Moles / Molarity = 0.020 moles / 1.0 M = 0.020 L or 20.0 mL.

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Most popular questions from this chapter

$$\mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ Increasing the temperature of the above reaction will increase the rate of reaction. Which of the following is NOT a reason that increased temperature increases reaction rate? (A) The reactants will be more likely to overcome the activation energy. (B) The number of collisions between reactant molecules will increase. (C) A greater distribution of reactant molecules will have high velocities. (D) Alternate reaction pathways become available at higher temperatures.

A rigid, sealed 12.00 \(\mathrm{L}\) container is filled with 10.00 \(\mathrm{g}\) each of three different gases: \(\mathrm{CO}_{2}, \mathrm{NO},\) and \(\mathrm{NH}_{3}\) . The temperature of the gases is held constant \(35.0^{\circ} \mathrm{C} .\) Assume ideal behavior for all gases. (a) (i) What is the mole fraction of each gas? (ii) What is the partial pressure of each gas? (b) Out of the three gases, molecules of which gas will have the highest velocity? Why? (c) Name one circumstance in which the gases might deviate from ideal behavior, and clearly explain the reason for the deviation.

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) $$2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}$$ As the reaction progresses at a constant temperature of 1000 K, how does the value for the Gibbs free energy constant for the reaction change? (A) It stays constant. (B) It increases exponentially. (C) It increases linearly. (D) It decreases exponentially.

Aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\) is a weak base with \(K_{\mathrm{b}}=3.8 \times 10^{-10}\) (a) Write out the reaction that occurs when aniline reacts with water. (b) (i) What is the concentration of each species at equilibrium in a solution of 0.25\(M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} ?\) (ii) What is the pH value for the solution in (i)?

What is the general relationship between temperature and entropy for diatomic gases? (A) They are completely independent of each other; temperature has no effect on entropy. (B) There is a direct relationship, because at higher temperatures there is an increase in energy dispersal. (C) There is an inverse relationship, because at higher temperatures substances are more likely to be in a gaseous state. (D) It depends on the specific gas and the strength of the intermolecular forces between individual molecules.
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