Chapter 7: Problem 61
The \(\mathrm{pH}\) of the solution containing \(10 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) \(\mathrm{NaOH}\) and \(10 \mathrm{ml}\) of \(0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) would be (1) zero (2) 1 (3) \(>7\) (4) 7
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid-Base Titration
Let's start by understanding acid-base titration. It is a laboratory method used to determine the concentration of an acid or base in a solution by adding a known amount of titrant (acid or base) until the reaction reaches its end-point, typically indicated by a color change in an indicator.
During the titration process, an acid and a base react to form water and a salt, a process known as neutralization.
By using data from the titration, we can calculate the concentration of one of the solutions using the equation: \ \text{{N}}_1 \text{{V}}_1 = \text{{N}}_2 \text{{V}}_2, where \ \( N_1 \) and \( V_1 \) are the normality and volume of the titrant, and \( N_2 \) and \( V_2 \) are the normality and volume of the analyte.
During the titration process, an acid and a base react to form water and a salt, a process known as neutralization.
By using data from the titration, we can calculate the concentration of one of the solutions using the equation: \ \text{{N}}_1 \text{{V}}_1 = \text{{N}}_2 \text{{V}}_2, where \ \( N_1 \) and \( V_1 \) are the normality and volume of the titrant, and \( N_2 \) and \( V_2 \) are the normality and volume of the analyte.
Neutralization Reaction
A neutralization reaction occurs when an acid reacts with a base to form water and a salt.
In the exercise, \( \text{{H}}_2 \text{{SO}}_4 \) (sulfuric acid) and \( \text{{NaOH}} \) (sodium hydroxide) are involved in a neutralization reaction.
This reaction can be written as: \( \text{{H}}_2 \text{{SO}}_4 + 2 \text{{NaOH}} \rightarrow 2 \text{{H}}_2 \text{{O}} + \text{{Na}}_2 \text{{SO}}_4 \).
Here, 1 mole of \( \text{{H}}_2 \text{{SO}}_4 \) reacts with 2 moles of \( \text{{NaOH}} \) to produce water and sodium sulfate.
Each reactant is used up completely to form neutral water and a salt, indicating a stoichiometric completion where all equivalent acid and base are consumed.
In the exercise, \( \text{{H}}_2 \text{{SO}}_4 \) (sulfuric acid) and \( \text{{NaOH}} \) (sodium hydroxide) are involved in a neutralization reaction.
This reaction can be written as: \( \text{{H}}_2 \text{{SO}}_4 + 2 \text{{NaOH}} \rightarrow 2 \text{{H}}_2 \text{{O}} + \text{{Na}}_2 \text{{SO}}_4 \).
Here, 1 mole of \( \text{{H}}_2 \text{{SO}}_4 \) reacts with 2 moles of \( \text{{NaOH}} \) to produce water and sodium sulfate.
Each reactant is used up completely to form neutral water and a salt, indicating a stoichiometric completion where all equivalent acid and base are consumed.
pH of Solutions
The pH of a solution measures its acidity or basicity.
It is calculated using the formula: \( \text{{pH}} = -\log [\text{{H}}^+] \), where \( [\text{{H}}^+] \) is the concentration of hydrogen ions.
In neutralization, if equal amounts of a strong acid and strong base mix, the \( \text{{H}}^+ \) and \( \text{{OH}}^- \) ions completely neutralize each other, leading to a pH of 7.
For the exercise provided, after calculating the milliequivalents of both \( \text{{NaOH}} \) and \( \text{{H}}_2 \text{{SO}}_4 \), the solution is neutral with a pH of 7 since they neutralize each other completely.
It is calculated using the formula: \( \text{{pH}} = -\log [\text{{H}}^+] \), where \( [\text{{H}}^+] \) is the concentration of hydrogen ions.
In neutralization, if equal amounts of a strong acid and strong base mix, the \( \text{{H}}^+ \) and \( \text{{OH}}^- \) ions completely neutralize each other, leading to a pH of 7.
For the exercise provided, after calculating the milliequivalents of both \( \text{{NaOH}} \) and \( \text{{H}}_2 \text{{SO}}_4 \), the solution is neutral with a pH of 7 since they neutralize each other completely.
Normality and Molarity
Normality (N) and molarity (M) are both measures of concentration.
Molarity is the number of moles of solute per liter of solution: \( M = \frac{{moles\text{{ of solute}}}}{{liters \text{{ of solution}}}} \).
Normality is the gram equivalent weight of solute per liter of solution and takes into account the reactivity of the solute: \( N = \frac{{equivalents \text{{ of solute}}}}{{liters \text{{ of solution}}}} \).
For acids and bases, normality accounts for the number of protons or hydroxide ions a substance can donate or accept.
In the given exercise, we converted the molarity of \( \text{{H}}_2 \text{{SO}}_4 \) to normality, recognizing that it is a diprotic acid and thus its normality is twice its molarity.
Molarity is the number of moles of solute per liter of solution: \( M = \frac{{moles\text{{ of solute}}}}{{liters \text{{ of solution}}}} \).
Normality is the gram equivalent weight of solute per liter of solution and takes into account the reactivity of the solute: \( N = \frac{{equivalents \text{{ of solute}}}}{{liters \text{{ of solution}}}} \).
For acids and bases, normality accounts for the number of protons or hydroxide ions a substance can donate or accept.
In the given exercise, we converted the molarity of \( \text{{H}}_2 \text{{SO}}_4 \) to normality, recognizing that it is a diprotic acid and thus its normality is twice its molarity.
