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Chapter 8: Problem 44

If \(29.7 \mathrm{~mL}\) of a \(0.205 \mathrm{M} \mathrm{KOH}\) solution is required to neutralize \(25.0 \mathrm{~mL}\) of \(\mathrm{a} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) solution, what is the molarity of the acetic acid solution? \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
The molarity of the acetic acid solution is 0.244 M.

Step by step solution

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01

Write the balanced equation

Refer to the balanced chemical equation provided: \[ \text{HC}_2 \text{H}_3 \text{O}_2 (\text{aq}) + \text{KOH} (\text{aq}) \rightarrow \text{KC}_2 \text{H}_3 \text{O}_2 (\text{aq}) + \text{H}_2 \text{O} (\text{l}) \] This equation will help in understanding the mole ratio between the reactants.
02

Calculate moles of KOH

Use the molarity equation, \(n = M \times V\), to find the moles of KOH. The volume should be converted to liters: \[ V_{\text{KOH}} = 29.7 \text{ mL} = 0.0297 \text{ L} \] \[ n_{\text{KOH}} = 0.205 \text{ M} \times 0.0297 \text{ L} \] \[ n_{\text{KOH}} = 0.00608985 \text{ moles} \].
03

Determine moles of acetic acid

From the balanced chemical equation, the mole ratio of KOH to HC\textsubscript{2}H\textsubscript{3}O\textsubscript{2} is 1:1. Therefore, the moles of acetic acid are equal to the moles of KOH: \[ n_{\text{HC}_2 \text{H}_3 \text{O}_2} = n_{\text{KOH}} = 0.00608985 \text{ moles} \]
04

Calculate the molarity of the acetic acid solution

Again use the molarity equation, \(M = \frac{n}{V}\), where the volume should be in liters: \[ V_{\text{HC}_2 \text{H}_3 \text{O}_2} = 25.0 \text{ mL} = 0.0250 \text{ L} \] \[ M_{\text{HC}_2 \text{H}_3 \text{O}_2} = \frac{0.00608985 \text{ moles}}{0.0250 \text{ L}} \] \[ M_{\text{HC}_2 \text{H}_3 \text{O}_2} = 0.2436 \text{ M} \]

Key Concepts

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

Neutralization Reaction
In a neutralization reaction, an acid reacts with a base to produce a salt and water. The chemical equation involved in this exercise is:
\[ \text{HC}_2 \text{H}_3 \text{O}_2 (\text{aq}) + \text{KOH} (\text{aq}) \rightarrow \text{KC}_2 \text{H}_3 \text{O}_2 (\text{aq}) + \text{H}_2 \text{O} (\text{l}) \]
Here, acetic acid (\(\text{HC}_2\text{H}_3\text{O}_2\)) is the acid, and potassium hydroxide (\(\text{KOH}\)) is the base. When they react, they form potassium acetate (\(\text{KC}_2\text{H}_3\text{O}_2\)) and water (\(\text{H}_2\text{O}\)). Understanding this balanced reaction helps in determining the relationship between the moles of reactants and products.
For every mole of acetic acid, one mole of potassium hydroxide is required to complete the reaction. Thus, the mole ratio between acetic acid and potassium hydroxide is always 1:1. This simplifies the calculations, as the moles of acid and base will be equal at neutralization. This foundational concept of neutralization reaction is crucial for solving problems related to acid-base titrations.
Mole Concept
The mole concept is fundamental in chemistry and is used to quantify the amount of a substance. In this exercise, it helps us determine the amount of reactants needed for the neutralization reaction.
Moles can be calculated using the formula: \( n = M \times V \)
where:
  • \(n\) is the number of moles
  • \(M\) is the molarity (moles per liter)
  • \(V\) is the volume in liters

To find the moles of \(\text{KOH}\), we use:
\[ V_{\text{KOH}} = 29.7 \text{ mL} = 0.0297 \text{ L} \ n_{\text{KOH}} = 0.205 \text{ M} \times 0.0297 \text{ L} \ n_{\text{KOH}} = 0.00608985 \text{ moles} \]
Remember to always convert the volume to liters before calculating. The mole ratio from the balanced equation tells us that the moles of \(\text{HC}_2 \text{H}_3 \text{O}_2\) will be the same as the moles of \(\text{KOH}\):
\[ n_{\text{HC}_2 \text{H}_3 \text{O}_2} = 0.00608985 \text{ moles} \]
Understanding the mole concept allows you to bridge the relationship between mass, volume, and the quantity of substances involved in chemical reactions.
Molarity
Molarity (\(M\)), also called the molar concentration, is the number of moles of a solute per liter of solution. It is a key concept in titration problems as it relates the moles of a substance to the volume of the solution.
The formula to calculate molarity is: \[ M = \frac{n}{V} \]
where:
  • \(M\) is the molarity
  • \(n\) is the number of moles
  • \(V\) is the volume in liters

In this exercise, we need to find the molarity of acetic acid. We already have the moles of acetic acid (\(n \approx 0.00608985\text{ moles}\)) and the volume of the \(\text{HC}_2\text{H}_3\text{O}_2\) solution in liters:
\[ V_{\text{HC}_2 \text{H}_3 \text{O}_2} = 25.0 \text{ mL} = 0.0250 \text{ L} \]
Plugging these values into the molarity formula gives:
\[ M_{\text{HC}_2\text{H}_3\text{O}_2} = \frac{0.00608985 \text{ moles}}{0.0250 \text{ L}} \approx 0.2436 \text{ M} \]
This means the molarity of the acetic acid solution is \(0.2436\text{ M}\), which quantifies its concentration. Molarity is useful in determining how strong a solution is and in predicting how it will react with other substances.

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Most popular questions from this chapter

Complete the following table: $$ \begin{array}{|l|l|} \hline \text { Acid } & \text { Conjugate Base } \\ \hline \mathrm{H}_{2} \mathrm{O} & \\ \hline & \mathrm{CN}^{-} \\ \hline \mathrm{HNO}_{2} & \\ \hline & \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \\ \hline \end{array} $$

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a solution with the following \(\mathrm{pH}\) values? a. \(3.0\) b. \(6.00\) c. \(8.0\) d. \(11.0\) e. \(9.20\)

Complete and balance the equation for each of the following reactions: a. \(\mathrm{ZnCO}_{3}(s)+\operatorname{HBr}(a q) \longrightarrow\) b. \(\mathrm{Zn}(s)+\mathrm{HCl}(a q) \longrightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaHCO}_{3}(s) \longrightarrow\) d. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow\)

Calculate the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) of each aqueous solution with the following \(\left[\mathrm{OH}^{-}\right]:\) a. baking soda, \(1.0 \times 10^{-6} \mathrm{M}\) b. orange juice, \(5.0 \times 10^{-11} \mathrm{M}\) c. milk, \(2.0 \times 10^{-8} \mathrm{M}\) d. bleach, \(2.1 \times 10^{-3} \mathrm{M}\)

Calculate the \(\mathrm{pH}\) of each solution given the following \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) or \(\left[\mathrm{OH}^{-}\right]\) values: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1 \times 10^{-8} \mathrm{M}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-6} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=4 \times 10^{-2} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=8 \times 10^{-3} \mathrm{M}\) e. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=4.7 \times 10^{-2} \mathrm{M}\) f. \(\left[\mathrm{OH}^{-}\right]=3.9 \times 10^{-6} \mathrm{M}\)
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