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Chapter 1: Problem 3

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point? (A) The change would reduce the amount as the acid now fully dissociates. (B) The change would reduce the amount because the base will be more strongly attracted to the acid. (C) The change would increase the amount because the reaction will now go to completion instead of equilibrium. (D) Changing the strength of the acid will not change the volume needed to reach equivalence.

Short Answer

Expert verified
The correct answer is (D) Changing the strength of the acid will not change the volume needed to reach equivalence.

Step by step solution

01

Calculate moles of NaOH

First, calculate moles of NaOH using the formula: moles = molarity * volume(liters) = 1.0 M * 0.020 L = 0.020 moles.
02

Calculate moles of HCO2H required

Since formic acid and NaOH react in a 1:1 ratio, the moles of HCO2H required to reach the equivalence point is equal to the moles of NaOH, which is 0.020 moles.
03

Calculate volume of HCO2H required

Next, calculate the volume of HCO2H required to provide these moles using the formula: volume(liters) = moles / molarity = 0.020 moles / 2.0 M = 0.010 L = 10.0 mL.
04

Equivalence point with HCl

Now, consider what happens if HCO2H is replaced with HCl. Since HCl is a strong acid, it fully dissociates, but it should still react with NaOH in a 1:1 ratio. Therefore, if the concentration of HCl is kept the same, the volume required to reach the equivalence point will remain the same.

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Most popular questions from this chapter

For a reaction involving nitrogen monoxide inside a sealed flask, the value for the reaction quotient \((Q)\) was found to be \(1.1 \times 10^{2}\) at a given point. If, after this point, the amount of NO gas in the flask increased, which reaction is most likely taking place in the flask? (A) \(\operatorname{NOBr}(g) \rightarrow \operatorname{NO}(g)+1 / \operatorname{Br}_{2}(g) \quad K_{\mathrm{C}}=3.4 \times 10^{-2}\) (B) \(2 \mathrm{NOCl}(g) \mapsto 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K_{\mathrm{c}}=1.6 \times 10^{-5}\) (C) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{\mathrm{c}}=4.0 \times 10^{6}\) (D) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.2 \times 10^{2}\)

Nitrogen’s electronegativity value is between those of phosphorus and oxygen. Which of the following correctly describes the relationship between the three values? (A) The value for nitrogen is less than that of phosphorus because nitrogen is larger, but greater than that of oxygen because nitrogen has a greater effective nuclear charge. (B) The value for nitrogen is less than that of phosphorus because nitrogen has fewer protons, but greater than that of oxygen because nitrogen has fewer valence electrons. (C) The value for nitrogen is greater than that of phosphorus because nitrogen has fewer electrons, but less than that of oxygen because nitrogen is smaller. (D) The value for nitrogen is greater than that of phosphorus because nitrogen is smaller, but less than that of oxygen because nitrogen has a smaller effective nuclear charge.

\(\begin{array}{ll}{\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}} & {E^{\circ}=+0.3 \mathrm{V}} \\ {\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}} & {E^{\circ}=-0.8 \mathrm{V}} \\ {\mathrm{Mn}^{2+}+2 e^{-} \rightarrow \mathrm{Mn}} & {E^{\circ}=-1.2 \mathrm{V}}\end{array}\) Based on the reduction potentials given above, which of the following reactions will be favored? (A) \(\mathrm{Mn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Mn}+\mathrm{Cu}^{2+}\) (B) \(\mathrm{Mn}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Mn}+\mathrm{Zn}^{2+}\) (C) \(\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}\) (D) \(\mathrm{Zn}^{2+}+\mathrm{Mn} \rightarrow \mathrm{Zn}+\mathrm{Mn}^{2+}\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

Which of the substances would be soluble in water? (A) Ethylene glycol only, because it has the longest bond lengths (B) Acetone only, because it is the most symmetrical (C) Ethanol and ethylene glycol only, because of their hydroxyl (-OH) (D) All three substances would be soluble in water due to their permanent dipoles.
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