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Hydrogen ion concentration in \(\mathrm{mol} / \mathrm{L}\) in solution of \(\mathrm{pH} 5.4\) will be: (a) \(3.88 \times 10^{6}\) (b) \(3.98 \times 10^{-6}\) (c) \(3.98 \times 10^{8}\) (d) \(3.68 \times 10^{-6}\)

Short Answer

Expert verified
The hydrogen ion concentration is \(3.98 \times 10^{-6}\, \mathrm{mol/L}\), option (b).

Step by step solution

01

Understanding pH Definition

The pH of a solution is a measure of its hydrogen ion concentration. It is calculated using the formula: \( pH = -\log[H^+] \), where \([H^+]\) is the hydrogen ion concentration in moles per liter.
02

Rearranging the pH Formula

To find the hydrogen ion concentration, rearrange the formula to isolate \([H^+]\): \([H^+] = 10^{-pH}\). This equation will allow us to calculate the hydrogen ion concentration from the given pH value.
03

Substituting the Known pH Value

Substitute the given pH value (5.4) into the rearranged formula: \([H^+] = 10^{-5.4}\). This substitution helps to obtain the numerical value for the hydrogen ion concentration.
04

Calculating the Hydrogen Ion Concentration

Using a calculator, evaluate \(10^{-5.4}\). The calculation yields \([H^+] \approx 3.98 \times 10^{-6}\). This value represents the hydrogen ion concentration in \(\mathrm{mol/L}\).
05

Selecting the Correct Answer

Compare the calculated hydrogen ion concentration \(3.98 \times 10^{-6}\) with the provided answer choices. The correct matching option is (b) \(3.98 \times 10^{-6}\).

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Key Concepts

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

Hydrogen Ion Concentration
The hydrogen ion concentration helps us determine how acidic or basic a solution is. A higher concentration of hydrogen ions indicates a more acidic solution. The concentration is typically measured in moles per liter (mol/L).
Understanding the concentration of hydrogen ions is especially crucial in fields like chemistry and biology. Knowing the acidity of a solution can affect chemical reactions, the environment's balance, and even biological processes.
  • For instance, a high hydrogen ion concentration can cause metals to corrode quicker.
  • In biological systems, enzymes work optimally at specific hydrogen ion concentrations.
In everyday life, this concept plays a vital role in products like medicines and cleaning agents. They rely on specific hydrogen ion concentrations to be effective and safe to use.
Acid-Base Chemistry
Acid-base chemistry focuses on acids and bases and their reactions. An acid has a higher concentration of hydrogen ions and releases them when dissolved in water. Bases, on the other hand, accept these hydrogen ions, often resulting in the formation of water.
This dual nature of acids and bases creates a balance known as acid-base equilibrium. It plays a significant role in various chemical reactions. A practical example is the neutralization reaction, where an acid reacts with a base to form water and salt.
Acid-base concepts are essential not only in laboratory settings but also in real-world applications.
  • Baking relies on acid-base reactions for leavening.
  • Acidic soil can affect how plants absorb nutrients.
The understanding of acids and bases is fundamental in crafting solutions, whether for scientific experiments or cooking recipes.
Logarithms in Chemistry
In chemistry, logarithms help simplify calculations involving very large or very small numbers, like hydrogen ion concentrations. The pH scale uses logarithmic calculations, making it easier to express these concentrations.
The relationship is expressed as: \[ pH = -\log[H^+] \] This equation shows how logarithms transform exponential values into a convenient scale for everyday use. A small change in pH value represents a tenfold change in hydrogen ion concentration, illustrating how sensitive this scale is.
Utilizing logarithms also helps chemists understand reaction rates and equilibria.
  • They aid in predicting the direction of a reaction.
  • Logarithms also simplify expression of the energy changes during a reaction.
Overall, understanding logarithms enhances our ability to manipulate and interpret chemical data more effectively.

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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.

Which of the following change will shift the reaction in forward direction: \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g})\) Take \(\Delta \mathrm{H}^{\circ}=+150 \mathrm{~kJ}\) (a) Increase in concentration of I (b) Increase in total pressure (c) Decrease in concentration of \(\mathrm{I}_{2}\) (d) Increase in temperature

The equilibrium constant for the reaction: \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}\). The value of \(\mathrm{K}_{c}\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

The equilibrium between water and its vapour, in an open vessel: (a) Can be achieved (b) Depends upon pressure (c) Cannot be achieved (d) Depends upon temperature

For the reaction \(\begin{aligned}&\mathrm{PQ}_{2} \rightleftharpoons \mathrm{PQ}+\mathrm{Q} \text { the degree of } \\\&' \mathrm{~g} & \mathrm{~g}\end{aligned} \mathrm{~g}\). dissociation \((\alpha)\) of \(\mathrm{PQ}_{2}\) can be related with \(\mathrm{P}\) or \(\mathrm{V}\) as: (a) \(\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}\) (b) \(\alpha \propto \frac{1}{\mathrm{P}}\) (c) \(\alpha \propto \frac{1}{\sqrt{\mathrm{V}}}\) (d) \(\alpha \propto \mathrm{V}\)

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