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Chapter 15: Problem 31

How many milliliters of \(0.100 \mathrm{M}\) potassium hydroxide are required to completely neutralize \(10.0 \mathrm{~mL}\) of \(0.165 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3} ?\) $$ \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+2 \mathrm{KOH}(a q) \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Short Answer

Expert verified
33.0 mL of 0.100 M KOH is required.

Step by step solution

01

Identify Chemical Reaction and Ratios

The reaction between potassium hydroxide \( \text{KOH} \) and sulfurous acid \( \text{H}_2\text{SO}_3 \) is given by:\[\text{H}_2\text{SO}_3(aq) + 2 \text{KOH}(aq) \rightarrow \text{K}_2\text{SO}_3(aq) + 2 \text{H}_2\text{O}(l)\]This balanced equation indicates that 2 moles of \( \text{KOH} \) are required to neutralize 1 mole of \( \text{H}_2\text{SO}_3 \).
02

Calculate Moles of \( \text{H}_2\text{SO}_3 \)

First, calculate the moles of \( \text{H}_2\text{SO}_3 \) in the solution. The formula for moles is:\[\text{moles} = \text{Molarity} \times \text{Volume (L)}\]Convert the volume from milliliters to liters and compute:\[10.0 \text{ mL} = 0.0100 \text{ L}\]\[\text{moles of } \text{H}_2\text{SO}_3 = 0.165 \text{ M} \times 0.0100 \text{ L} = 0.00165 \text{ moles}\]
03

Use the Stoichiometric Relationship

According to the balanced equation, 2 moles of \( \text{KOH} \) are required for every mole of \( \text{H}_2\text{SO}_3 \). Therefore, calculate the needed moles of \( \text{KOH} \):\[\text{moles of } \text{KOH} = 2 \times 0.00165 = 0.00330 \text{ moles}\]
04

Calculate Volume of \( \text{KOH} \) Solution Needed

To find the volume of \( \text{KOH} \) solution needed, use the equation for moles:\[\text{moles} = \text{Molarity} \times \text{Volume (L)}\]Rearrange to find the volume:\[\text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}}\]Substitute the known values:\[\text{Volume of } \text{KOH} = \frac{0.00330 \text{ moles}}{0.100 \text{ M}} = 0.0330 \text{ L}\]Convert liters to milliliters:\[0.0330 \text{ L} = 33.0 \text{ mL}\]
05

Final Answer

Thus, the volume of \(0.100 \text{ M} \) potassium hydroxide required to completely neutralize \(10.0 \text{ mL} \) of \(0.165 \text{ M} \) sulfurous acid is \(33.0 \text{ mL} \).

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the principle that matter is conserved during a chemical process. To perform stoichiometric calculations:
  • First, write and balance the chemical equation. This ensures that the number of atoms for each element is the same on both sides of the reaction.
  • Identify the mole ratio of the substances involved. This ratio comes from the coefficients in the balanced chemical equation.
  • Use this ratio to calculate unknown quantities of reactants or products, like moles or volume needed for the reaction to proceed.
The balanced equation \[\text{H}_2\text{SO}_3(aq) + 2 \text{KOH}(aq) \rightarrow \text{K}_2\text{SO}_3(aq) + 2 \text{H}_2\text{O}(l)\]shows a 1:2 mole ratio between \( \text{H}_2\text{SO}_3 \) and \( \text{KOH} \), meaning two moles of \( \text{KOH} \) are necessary to fully react with one mole of \( \text{H}_2\text{SO}_3 \). With stoichiometry, we can precisely compute how much of each substance is needed or produced in a reaction.
Neutralization Reaction
Neutralization reactions occur when an acid and a base react to form water and a salt. These reactions are crucial in chemistry and daily life, such as in antacid medication. The neutralization of sulfurous acid \( \text{H}_2\text{SO}_3 \) with potassium hydroxide \( \text{KOH} \) is a classic example. In this particular reaction:
  • \( \text{H}_2\text{SO}_3 \), a diprotic acid, donates two protons \( \text{H}^+ \).
  • Each \( \text{KOH} \) molecule provides one hydroxide ion \( \text{OH}^- \).
  • To neutralize the acid completely, two hydroxide ions are required for each acid molecule, forming two water molecules \( \text{H}_2\text{O} \) and potassium sulfite \( \text{K}_2\text{SO}_3 \).
This reaction exemplifies a balanced neutralization, where the acid and base quantities are tuned so that neither is in excess once the reaction concludes.
Molarity Calculation
Molarity is a measure of the concentration of a solute in a solution, expressed in moles per liter (M). To calculate molarity:
  • Determine the number of moles of solute, using the relation \(\text{moles} = \text{Molarity} \times \text{Volume (L)}\).
  • Rearrange the formula to solve for unknown variables such as molarity or volume, depending on the given information.
  • For precise results, convert all measurements to appropriate units, such as converting milliliters to liters by dividing by 1000.
In the original exercise, calculate the moles of \( \text{H}_2\text{SO}_3 \) required: Convert \(10.0 \text{ mL} \) to liters:\[10.0 \text{ mL} = 0.0100 \text{ L}\]Calculate moles:\[\text{moles} = 0.165 \text{ M} \times 0.0100 \text{ L} = 0.00165 \text{ moles}\]Using the molarity formula, we then determined the volume required to neutralize completely. Understanding these calculations enables precise measurement of reactant and product concentrations, crucial for laboratory and industrial applications.

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Most popular questions from this chapter

Determine the acid and base that were neutralized to produce each of the following salts: (a) sodium fluoride, \(\operatorname{NaF}(a q)\) (b) magnesium iodide, \(\operatorname{MgI}_{2}(a q)\) (c) calcium nitrate, \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (d) lithium carbonate, \(\mathrm{Li}_{2} \mathrm{CO}_{3}(a q)\)

Ascorbic acid, abbreviated HAsc, is the chemical name for vitamin C. If \(30.95 \mathrm{~mL}\) of \(0.176 \mathrm{M} \mathrm{NaOH}\) neutralizes \(0.959 \mathrm{~g}\) of acid, what is the molar mass of vitamin C? $$ \operatorname{HAsc}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaAsc}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$

Complete and balance the following Bronsted-Lowry neutralization reactions: (a) \(\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \longrightarrow\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow\)

Calculate the \(\mathrm{pH}\) of each of the following given the molar hydroxide ion concentration: (a) \(\left[\mathrm{OH}^{-}\right]=0.11 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=0.00055 \mathrm{M}\)

Classify each of the following aqueous solutions as a strong or weak electrolyte: (a) \(\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{3}(a q)\) (b) \(\mathrm{AlPO}_{4}(\mathrm{~s})\) (c) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})\) (d) \(\mathrm{CdSO}_{4}(a q)\)
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