Chapter 16: Problem 125
The equivalent mass of phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) in the reaction, \(\mathrm{NaOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{NaH}_{2} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) is (a) 98 (b) 89 (c) 49 (d) 58
Short Answer
Expert verified
The equivalent mass of \(\mathrm{H}_3\mathrm{PO}_4\) is 98 g/equivalent.
Step by step solution
01
Determine Molar Mass of \(\mathrm{H}_3 \mathrm{PO}_4\)
Calculate the molar mass of \(\mathrm{H}_3 \mathrm{PO}_4\) by adding the atomic masses of its constituent elements: three hydrogen atoms, one phosphorus atom, and four oxygen atoms. Using atomic masses, we have \(3 \times 1 + 31 + 4 \times 16 = 98 \, \text{grams/mole}.\) So, the molar mass of \(\mathrm{H}_3 \mathrm{PO}_4\) is 98 g/mol.
02
Identify the Reaction Type
In the reaction provided, \(\mathrm{NaOH}\) reacts with \(\mathrm{H}_3 \mathrm{PO}_4\) to form \(\mathrm{NaH}_2 \mathrm{PO}_4\) and \(\mathrm{H}_2 \mathrm{O}\). This is an acid-base neutralization reaction where \(\mathrm{H}_3 \mathrm{PO}_4\) donates one proton (\(\mathrm{H}^+\)).
03
Determine Equivalent Mass
The equivalent mass of an acid is calculated as the molar mass divided by the number of replaceable hydrogen ions (protons) per molecule. Since \(\mathrm{H}_3 \mathrm{PO}_4\) donates one proton in this reaction, its equivalent mass is \(\frac{98}{1} = 98\, \text{grams/equivalent}.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass Calculation
To understand molar mass calculation, let's first think of molar mass as the weight of one mole of a substance. It's like finding the total mass of a box of chocolates where each piece has a different weight. For phosphoric acid, \( \mathrm{H}_{3}\mathrm{PO}_{4} \), we add up the atomic masses of all the atoms present in one molecule:
- Hydrogen: 3 atoms, each with a mass of about 1 gram/mole.
- Phosphorus: 1 atom with a mass of about 31 grams/mole.
- Oxygen: 4 atoms, each with a mass of about 16 grams/mole.
Acid-Base Neutralization
Acid-base neutralization is a fundamental concept where an acid reacts with a base to form water and a salt. Imagine a seesaw balancing act where one side gives and one side takes until it evens out. In our exercise, \( \mathrm{NaOH} \), a base, reacts with \( \mathrm{H}_{3}\mathrm{PO}_{4} \), an acid, to yield \( \mathrm{NaH}_{2}\mathrm{PO}_{4} \) and water.
This forms the foundation of neutralization reactions.
It typically involves:
This forms the foundation of neutralization reactions.
It typically involves:
- A transfer of protons from acid to base.
- The production of water, which is neither acidic nor basic.
Proton Donation
In the context of acid-base chemistry, proton donation refers to the process where an acid provides protons (\( \mathrm{H}^+ \)) to a base. Think of the acid as a friendly neighbor giving away apples, which in this case are protons. \( \mathrm{H}_{3}\mathrm{PO}_{4} \) donates a proton to \( \mathrm{NaOH} \).
This single proton donation is key to determining equivalent mass and reaction outcome.
Key points about proton donation include:
This single proton donation is key to determining equivalent mass and reaction outcome.
Key points about proton donation include:
- Acids are defined by their ability to donate protons.
- The number of protons an acid can donate influences reactions and equivalence.
Chemical Reactions
Chemical reactions are processes where substances, the reactants, are transformed into different substances, the products. It’s like following a recipe where each ingredient mixed a certain way results in a new dish. In our given reaction, \( \mathrm{NaOH} \) and \( \mathrm{H}_{3}\mathrm{PO}_{4} \) are reactants that form \( \mathrm{NaH}_{2}\mathrm{PO}_{4} \) and \( \mathrm{H}_{2}\mathrm{O} \) as products.
- Reactions always involve bond breaking and forming.
- Each reaction has specific reactants and products determined by conditions like temperature or concentration.
Stoichiometry
Stoichiometry is the mathematics behind chemical reactions. It's like a blueprint that tells you exactly how much of each ingredient is needed to craft a new compound. Using stoichiometry, we balance chemical equations and determine amounts of products and reactants.
- It involves using balanced equations to find mole ratios.
- Helps predict how much product forms or how much reactant is needed.