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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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Most popular questions from this chapter

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 \(1000 \mathrm{K} :\) \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}\) (A) \(K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{S}\right]^{2}}\) (B) \(K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2} \mathrm{S}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{S}_{2}\right]}\) (C) \(K_{\mathrm{c}}=\frac{2\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]}{2\left[\mathrm{H}_{2} \mathrm{S}\right]}\) (D) \(K_{\mathrm{c}}=\frac{2\left[\mathrm{H}_{2} \mathrm{S}\right]}{2\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]}\)

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