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A centinormal solution of a monobasic acid is \(100 \%\) ionized. Its \(\mathrm{pH}\) is: (a) 2 (b) 4 (c) 3 (d) 1

Short Answer

Expert verified
The pH is 2 (option a).

Step by step solution

01

Understand the Problem

We have a centinormal (0.01 N) solution of a monobasic acid that is completely ionized, and we need to find its pH.
02

Recognize the Conditions

Since the acid is monobasic and 100% ionized, it completely dissociates in water. Thus, the concentration of hydrogen ions \([H^+]\) in the solution is equal to the concentration of the acid, 0.01 mol/L.
03

Use the pH Formula

The \( ext{pH}\) is calculated using the formula: \[\text{pH} = -\log [H^+]\] where \([H^+]\) is the concentration of hydrogen ions.
04

Calculate the pH

Substitute the given hydrogen ion concentration into the formula: \([H^+] = 0.01 = 10^{-2}\). Thus, \[\text{pH} = -\log(10^{-2}) = 2.\]
05

Compare with Options

The calculated pH value is 2. Therefore, match it with the given options: (a) 2, (b) 4, (c) 3, (d) 1. The correct answer is (a) 2.

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Key Concepts

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

Understanding Centinormal Solution
A centinormal solution is essentially a solution with a normality of 0.01 N. In simpler terms, the concentration of the solution is 0.01 equivalents per liter. This is a part of what is known as molarity, but specifically focused on equivalent reactive capacity.
The prefix 'centi' signifies one hundredth, which means that centinormal (cN) is equivalent to one hundredth of a normal solution (1 N).

Centinormal solutions are often used in titration experiments because they allow for more delicate measurements and are easier to work with due to their lower concentration.
  • If you start with a 1 normal solution, diluting it by a factor of 100 gives a centinormal solution.
  • Understanding centinormal solutions is essential for laboratory practices, where precision and accuracy are crucial.
In our problem, a centinormal solution of a monobasic acid makes the calculation straightforward because its properties aid smooth ionization understanding.
The Role of Monobasic Acid in pH Calculations
A monobasic acid is an acid that can donate only one proton (hydrogen ion) per molecule in an aqueous solution.
This is a very important concept because it greatly simplifies the way we calculate pH in solutions.

In a monobasic acid solution, you can directly assume that the concentration of the acid in moles is equal to the concentration of hydrogen ions \(H^+\) after ionization. This makes mathematical handling much easier.
  • Hydrochloric acid (HCl) is a common example of a monobasic acid.
  • In our exercise, the monobasic acid given is fully ionized, simplifying the calculation as the entire acid molecules dissociate.
This property allows you to directly use the concentration of the acid solution to find the pH, enabling accurate predictions of acidity in chemical reactions.
Ionization and Its Impact on Solution pH
Ionization is the process by which an atom or molecule acquires a negative or positive charge by gaining or losing electrons to form ions.
For acids in solution, this involves the dissociation of acid molecules into productive hydrogen ions \(H^+\) and other resultant ions.

In our scenario, the concept of 100% ionization means that every single molecule of the monobasic acid dissociates in the solution, forming hydrogen ions and leaving no un-ionized acid molecules.
  • This complete ionization ensures that the concentration of hydrogen ions is exactly the concentration of the acid, simplifying calculations.
  • 100% ionization is more often theoretical in nature, but it simplifies academic problems for foundational learning.
This characteristic of complete ionization in the exercise allows us to directly calculate pH using the formula \(-\log [H^+]\), quickly leading us to the precise value based on initial concentration.

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Most popular questions from this chapter

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves: (a) The equilibrium has been established in the solution (b) Radioactive sugar can displace non-radioactive sugar from its solution.

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole/litre: (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

The reaction \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) is in equilibrium. If the equilibrium concentration of \(\mathrm{PCI}_{3}\) (g) is doubled, then concentration of \(\mathrm{Cl}_{2}\) (g) would become: (a) \(1 / 2\) of its initial value (b) \(1 / 4\) of its initial value (c) Four times of its initial value (d) Two times of its initial value

In what manner will increase of pressure affect the following equation: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) Shift in the reverse direction (b) Shift in the forward direction (c) Increase in the yield of hydrogen (d) No effect

Phosphorous pentachloride dissociates as follows, in a closed reaction vessel. \(\mathrm{PCI}_{5}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\). If total pressure at equilibrium of the reaction mixture is \(\mathrm{P}\) and degree of dissociation of \(\mathrm{PC} 1_{5}\) is \(x\), the partial pressure of \(\mathrm{PCl}_{3}\) will be: (a) \(\left(\frac{x}{(x+1)}\right) \mathrm{P}\) (b) \(\left(\frac{2 x}{(x-1)}\right) \mathrm{P}\) (c) \(\left(\frac{x}{(x-1)}\right) \mathrm{P}\) (d) \(\left(\frac{x}{(1-x)}\right) \mathrm{P}\)

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