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Find the equivalent mass of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in the reaction $$ \begin{aligned} &\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{CaHPO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \\ &\text { (At. mass } \mathrm{H}=1, \mathrm{O}=16, \mathrm{P}=31, \mathrm{Ca}=40 \text { ) } \end{aligned} $$

Short Answer

Expert verified
The equivalent mass of H鈧働O鈧 is 49 g/equiv.

Step by step solution

01

Write the Balanced Chemical Reaction

The given chemical reaction is already balanced:\[ \mathrm{Ca(OH)_2} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{CaHPO}_4 + 2 \mathrm{H}_2\mathrm{O} \]
02

Determine the Role of H鈧働O鈧

H鈧働O鈧 is involved in the reaction by providing hydrogen ions (H鈦). In this reaction, H鈧働O鈧 acts as an acid and donates 2 moles of H鈦 ions to form water, as can be observed from the water part of the equation.
03

Calculate the Molar Mass of H鈧働O鈧

Use the atomic masses to calculate the molar mass of H鈧働O鈧: - H = 3 脳 1 = 3 - P = 31 脳 1 = 31 - O = 16 脳 4 = 64 Thus, the molar mass of H鈧働O鈧 is 3 + 31 + 64 = 98 g/mol.
04

Compute the Equivalent Mass of H鈧働O鈧

The equivalent mass of a compound in a reaction is its molar mass divided by the number of moles of hydrogen ions it donates or accepts in the reaction. Here, H鈧働O鈧 donates 2 moles of H鈦 ions:\[ \text{Equivalent Mass of } \mathrm{H}_3\mathrm{PO}_4 = \frac{98}{2} = 49 \, \text{g/equiv} \]

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Key Concepts

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

Molar Mass
Understanding molar mass is crucial when working with chemical reactions. The molar mass of a substance is the mass of one mole of its entities, determined by its atomic or molecular composition. This value is usually expressed in grams per mole (g/mol).
To find the molar mass of a compound like H鈧働O鈧, we sum the atomic masses of all the atoms within one molecule of this compound. For H鈧働O鈧:
  • Hydrogen (H) has an atomic mass of 1 g/mol, and there are 3 H atoms in the molecule: 3 脳 1 = 3 g/mol.
  • Phosphorus (P) has an atomic mass of 31 g/mol with one atom present, contributing: 31 g/mol.
  • Oxygen (O) has an atomic mass of 16 g/mol, and there are 4 O atoms, adding: 4 脳 16 = 64 g/mol.
Add these up: 3 + 31 + 64 = 98 g/mol. Thus, H鈧働O鈧 has a molar mass of 98 g/mol. Understanding how to calculate molar mass is essential for further calculations like equivalent mass in reactions.
Acid-Base Reaction
An acid-base reaction is a fundamental chemical reaction that involves the transfer of hydrogen ions (H鈦), often leading to the formation of water. Such reactions are central to many natural and industrial processes.
In the given reaction, Ca(OH)鈧 + H鈧働O鈧 鈫 CaHPO鈧 + 2 H鈧侽, H鈧働O鈧 acts as an acid, while Ca(OH)鈧 acts as a base. A base usually accepts hydrogen ions, which in this case, is facilitated by the hydroxide ions (OH鈦) from Ca(OH)鈧 forming water (H鈧侽).
These reactions are characterized by:
  • The acid donating hydrogen ions.
  • A base accepting these ions.
  • Formation of water or other neutral products.
Through this transfer, reactants shift into products, with H鈧働O鈧 donating protons to the hydroxide ions, showcasing the transformation of an acid-base interaction into water and a salt-like compound ( CaHPO鈧). Knowing the role of acids and bases is essential in predicting the products of such reactions.
Hydrogen Ion Donor
Hydrogen ion donor is a term frequently used to describe acids in acid-base reactions. An acid's capacity to donate hydrogen ions is a defining feature and directly impacts its chemical behavior.
In the reaction between Ca(OH)鈧 and H鈧働O鈧, H鈧働O鈧 acts as the hydrogen ion donor. As a triprotic acid, H鈧働O鈧 has three hydrogen ions available. However, in this particular reaction, it only donates 2 moles of hydrogen ions (H鈦).
The concept of a hydrogen ion donor helps explain:
  • How acids react with bases by transferring protons.
  • The calculation of equivalent mass, since it's based on the number of hydrogen ions donated.
  • Why certain acids can behave differently under varying conditions, depending on how many protons are released.
This understanding lets us accurately compute aspects like the equivalent mass, which is vital for quantified chemical calculations and reactions. In our example, the equivalent mass of H鈧働O鈧 is calculated as its molar mass divided by the number of hydrogen ions donated, illustrating its role as a hydrogen ion donor.

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Most popular questions from this chapter

\(1.26 \mathrm{~g}\) of a dibasic acid were dissolved in water and made up to \(200 \mathrm{~mL} .20 \mathrm{~mL}\) of this solution were completely neutralised by \(10 \mathrm{~mL}\) of \(\mathrm{N} / 5\) caustic soda solution. Calculate the equivalent mass and molecular mass of the acid.

(a) \(2 \mathrm{~g}\) of a metal carbonate were dissolved in \(50 \mathrm{~mL}\) of \(N\) HCl. \(100 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{NaOH}\) were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate. (b) How much water should be added to \(75 \mathrm{~mL}\) of \(3 \mathrm{NHCl}\) to make it a normal solution?

\(20 \mathrm{~mL} \mathrm{~N} / 2 \mathrm{HCl}, 60 \mathrm{~mL} \mathrm{~N} / 10 \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(150 \mathrm{~mL} \mathrm{~N} / 5 \mathrm{HNO}_{3}\) are mixed. Determine the normality of the mixture of the acids in solution.

An aqueous solution containing \(0.10 \mathrm{~g} \mathrm{KIO}_{3}\) (formula weight \(=214.0\) ) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated iodine consumed \(45.0 \mathrm{~mL}\) of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution. \(\quad\) [I.I.T. 1998] [Hint: \(2 \mathrm{KIO}_{3}+10 \mathrm{KI}+12 \mathrm{HCl} \longrightarrow 12 \mathrm{KCl}+6 \mathrm{I}_{2}+6 \mathrm{H}_{2} \mathrm{O}\) $$ \left[2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{NaI}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\right] \times 6 $$

How much water should be added to \(100 \mathrm{~mL}\) of \(2.5 \mathrm{NH}_{2} \mathrm{SO}_{4}\) to make it normal solution?

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