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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

Short Answer

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Step by step solution

01

- Identify the Normality of Acid and Base

The normality of NaOH is given as 0.1 N. The molarity of H₂SO₄ is given as 0.05 M. To convert this to normality, we recognize that H₂SO₄ is a diprotic acid, so its normality is twice its molarity: 0.05 M × 2 = 0.1 N.
02

- Calculate the Milliequivalents of Acid and Base

Calculate milliequivalents for NaOH: Volume × Normality = 10 mL × 0.1 N = 1 meq (milliequivalents)Calculate milliequivalents for H₂SO₄: Volume × Normality = 10 mL × 0.1 N = 1 meq
03

- Determine the Resultant Solution pH

Since both NaOH and Hâ‚‚SOâ‚„ have the same milliequivalents, they completely neutralize each other.
04

- Conclusion

After complete neutralization, the resulting solution is neutral, with a pH of 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.
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.
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.
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.

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Most popular questions from this chapter

Consider the following reactions (i) \(\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}\) (ii) \(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}\) (iii) \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}\) (iv) \(\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}\) Which pair of reactions proves that water is amphoteric? (1) \(i\) and ii (2) ii and iii (3) iii and iv (4) i and iii

\(\Lambda\) saturated solution of \(\mathrm{II}_{2} \mathrm{~S}\) in \(0.1 \mathrm{M} \mathrm{IICl}\) at \(25^{\circ} \mathrm{C}\) contains a \(\mathrm{S}^{2-}\) ion concentration of \(10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}\). The solubility products of some sulphidcs are: \(\mathrm{CuS}=10^{-44}\), \(\mathrm{FeS}=10^{-14} ; \mathrm{MnS}=10^{-15}\) and \(\mathrm{CdS}=10^{-25} .\) If \(0.01 \mathrm{M}\) solutions of these salts in \(1 \mathrm{M}\) IICl are saturated with \(\mathrm{H}_{2} \mathrm{~S}\), which of these will be precipitated? (1) All (2) All except MnS (3) All except MnS and FeS (4) Only CuS

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The decreasing order of the strength of bases \(\mathrm{OH}\), \(\mathrm{NH}_{2}, \mathrm{HC} \equiv \mathrm{C}\). and \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (1) \(\mathrm{CH}_{3} \mathrm{CH}_{2}>\mathrm{NH}_{2}>\mathrm{HC} \equiv \mathrm{C}>\mathrm{OH}\) (2) \(\mathrm{HC} \equiv \mathrm{C}>\mathrm{CH}_{3} \mathrm{CH}_{2}>\mathrm{NH}_{2}>\mathrm{OH}\) (3) \(\mathrm{OH}>\mathrm{NH}_{2}>\mathrm{HC} \equiv \mathrm{C}>\mathrm{CH}_{3}-\mathrm{CH}_{2}\) (4) \(\mathrm{NH}_{2}>\mathrm{HC} \equiv \mathrm{C}>\mathrm{OH}>\mathrm{CH}_{3}-\mathrm{CH}_{2}\)

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