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What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)

Short Answer

Expert verified
The required volume of \(0.0962 \, \mathrm{M} \, \mathrm{NaOH}\) needed to neutralize \(10.00 \, \mathrm{ml}\) of \(0.128 \, \mathrm{M} \, \mathrm{HCl}\) is \(13.3 \, \mathrm{ml}\).

Step by step solution

01

Determine the Amount of Moles of HCl

Calculate the moles of HCl using its volume and molarity. The molarity (\(M\)) could be defined as moles of solute per litre of solution (\(M=n/V\)). Therefore, to calculate the number of moles (\(n\)), we multiply the volume (\(V = 10.00 \, \mathrm{ml}\)) by its molarity (\(M = 0.128 \, \mathrm{M}\)). We convert the volume from ml to L (since molarity is in terms of L) by dividing by 1000. Then we solve, \(n=(10.00/1000) \times 0.128 = 0.00128 \, \mathrm{moles}\) of HCl.
02

Use Stoichiometry of the Balanced Equation

According to the balanced equation for the reaction, one mole of HCl reacts with one mole of NaOH. Consequently, the number of moles of NaOH required for neutralization equals the number of moles of HCl, which is \(0.00128 \, \mathrm{moles}\).
03

Compute the Volume of NaOH Needed

Now the calculated moles of NaOH can be used to find the volume needed by using rearranged molarity formula: \(V=n/M\). Here, \(n=0.00128 \, \mathrm{moles}\) and \(M=0.0962 \, \mathrm{M}\). After calculating, we find that \(V=0.00128/0.0962 = 0.0133 \, \mathrm{L} = 13.3 \, \mathrm{ml}\).

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry that refers to the concentration of a solute in a solution. It is expressed as moles of solute per litre of solution. Mathematically, it is given by the formula:
  • \[ M = \frac{n}{V} \]
where \(M\) is the molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of solution in litres. It's crucial for calculations involving reactions in solutions, as it helps determine the exact amount of substances needed.
To convert between mL and L, divide the volume in mL by 1000, because there are 1000 ml in a litre. This is particularly important when using the molarity formula, as the volume must be in litres to match the unit of molarity. Understanding molarity is essential for performing accurate chemical calculations and preparing solutions properly.
Neutralization
Neutralization is a chemical reaction where an acid and a base react to form a salt and water. This reaction is important in many everyday chemical processes, including antacid tablets reacting in the stomach to neutralize excess acid.
In a neutralization reaction, the acid donates H+ ions while the base donates OH- ions. These ions combine to form water, a process generally described by the equation:
  • \[ \text{H}^{+} + \text{OH}^{-} \rightarrow \text{H}_2\text{O} \]
Each neutralization reaction involves the stoichiometric ratio of the reactants, often shown in a balanced chemical equation. For example, hydrochloric acid (HCl) reacting with sodium hydroxide (NaOH) can be described as:
  • \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This tells us that one mole of HCl reacts with one mole of NaOH. Understanding these stoichiometric relationships is crucial in predicting the amounts of reactants needed to achieve neutralization.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products, and they are governed by specific stoichiometric ratios defined by balanced chemical equations. Understanding the concepts of reactants and products, along with how substances interact, is fundamental in chemistry.
When working with chemical reactions, it is important to observe the conservation of mass, indicating that the total mass of reactants equals the total mass of products. Equations must be balanced to reflect this.
There are various types of chemical reactions, but in the context of neutralization, acid-base reactions are key. These reactions often produce a salt and water, as exemplified by the reaction of HCl with NaOH. Balancing these equations requires knowing the exact quantities of substances that react, which can be determined using stoichiometry, allowing for precise calculations in chemistry labs and industrial processes alike. By mastering chemical reactions, you can predict outcomes in chemical experiments and understand the core principles of matter interaction.

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Most popular questions from this chapter

Balance these equations for disproportionation reactions. (a) \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{MnO}_{4}^{-}\) (basic solution) (b) \(\mathrm{P}_{4}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+\mathrm{PH}_{3}(\mathrm{g})\) (basic solution) (c) \(\mathrm{S}_{8}(\mathrm{s}) \longrightarrow \mathrm{S}^{2-}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (basic solution) (d) \(\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{AsO}_{4}^{3-}+\mathrm{SO}_{4}^{2-}\)

Every antacid contains one or more ingredients capable of reacting with excess stomach acid (HCl). The essential neutralization products are \(\mathrm{CO}_{2}\) and/ or \(\mathrm{H}_{2} \mathrm{O} .\) Write net ionic equations to represent the neutralizing action of the following popular antacids. (a) Alka-Seltzer (sodium bicarbonate) (b) Tums (calcium carbonate) (c) milk of magnesia (magnesium hydroxide) (d) Maalox (magnesium hydroxide, aluminum hydroxide) (e) Rolaids \(\left[\mathrm{NaAl}(\mathrm{OH})_{2} \mathrm{CO}_{3}\right]\)

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

Classify each of the following statements as true or false. (a) Barium chloride, \(\mathrm{BaCl}_{2^{\prime}}\) is a weak electrolyte in aqueous solution. (b) In the reaction \(\mathrm{H}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{OH}^{-}(\mathrm{aq}),\) water acts as both an acid and an oxidizing agent. (c) A precipitate forms when aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}),\) is treated with excess aqueous hydrochloric acid, HCl(aq). (d) Hydrofluoric acid, \(\overline{\mathrm{HF}}\), is a strong acid in water. (e) Compared with a 0.010 M solution of \(\mathrm{NaNO}_{3}\), a \(0.010 \mathrm{M}\) solution of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is a better conductor of electricity.

In your own words, define or explain the terms or symbols \((\mathrm{a}) \rightleftharpoons(\mathrm{b})[] ;(\mathrm{c})\) spectator ion; (d) weak acid.

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