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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. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Short Answer

Expert verified
(B) Basic; the only ion present at equilibrium is the conjugate base

Step by step solution

01

Understanding the given

In the given problem, a base \(1.0 M \mathrm{NaOH}\) is being titrated with an acid \(\mathrm{HCO}_{2} \mathrm{H}\) which has a given \(K_{a}\) value (indicating it is a weak acid). The base and acid have specified volumes (20.0 mL and we'll find out for acid).
02

Identifying the equivalence point

At the equivalence point, the number of moles of the base is equal to the number of moles of the acid.
03

Calculating the moles of base

Calculate the quantity in moles of NaOH using the formula n = M脳V, where n is the number of moles, M is molarity (concentration) and V is the volume. Here, M = 1.0 and the volume is 20 mL (we should convert this to liters = 0.020 L). We get: n = 1.0 脳 0.020 = 0.020 moles.
04

Determining the nature of the solution

In this case, as the acid gets completely titrated by the base, it will donate protons to form carbonic acid (H2CO3) leaving the conjugate base \(HC0_3 ^{-}\) in the solution. Because the acid is weak and doesn't completely ionize and the base is strong and ionizes completely (forming OH- ions), the solution will have excess OH- ions making it basic.

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Key Concepts

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

Equivalence Point
In an acid-base titration, the equivalence point is a critical moment. It is reached when the amount of acid equals the amount of base in a solution. This does not mean the solution is neutral. Whether it is acidic, basic, or neutral depends on the strengths of the reacting acid and base.
In the given exercise, the base is sodium hydroxide (NaOH), a strong base, and the acid is formic acid (HCO鈧侶), a weak acid. When they react, NaOH completely ionizes, while formic acid only partially ionizes because it is weak.
At the equivalence point, all of the NaOH has reacted with HCO鈧侶 to form water and the conjugate base of the weak acid, formate ion (HCO鈧傗伝). Since the conjugate base is left in the solution, the solution becomes basic after the equivalence point is reached.
Acid-Base Titration
An acid-base titration is a laboratory technique used to determine the concentration of an unknown acid or base. The process involves adding a titrant to a solution until the reaction reaches the equivalence point.
Key steps in titration include determining the point where the added titrant has completely reacted with the solution in the flask. For accuracy, it is essential to understand the nature of the acid and base involved.
  • For strong acids and strong bases, the equivalence point corresponds to a neutral pH.
  • For weak acids and strong bases, like formic acid with NaOH, the equivalence point tends to be in the basic range.
  • Indicators or pH meters can be used to detect the equivalence point visually or electronically.

In this exercise, we are dealing with formic acid and NaOH, where the resultant solution is basic even at the equivalence point.
Weak Acid
A weak acid is an acid that does not completely dissociate in water. Formic acid ( HCO鈧侶) is a prime example. Its dissociation into hydrogen ions (H鈦) and formate ions (HCO鈧傗伝) is partial, meaning only some of the molecules will ionize.
We measure the strength of an acid using the acid dissociation constant ( K鈧), which indicates how much of the acid ionizes. For formic acid, this value is small ( 1.8 脳 10鈦烩伌), confirming it is a weak acid.
In the context of titration, when a weak acid reacts with a strong base, the base consumes nearly all the hydrogen ions available from the weak acid. This leads to the formation of more of the weak acid's conjugate base, resulting in a basic solution at the equivalence point.
Strong Base
A strong base is fully ionized in water. Sodium hydroxide ( NaOH) is a classic strong base, which means it completely dissociates into sodium ions (Na鈦) and hydroxide ions (OH鈦).
Because strong bases fully dissociate, they effectively neutralize acids during a titration process. They react completely with the hydrogen ions from acids, including weak ones, transforming them into water and the salt of the weak acid.
In our exercise, the strong base, NaOH, fully reacts with formic acid, leaving formate ions, which create a basic environment. The strong nature of NaOH ensures that it can drive the ionization process to completion even when the reacting acid is weak.

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Most popular questions from this chapter

Nitrogen鈥檚 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.

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.

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