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A solution has hydrogen ion concentration \(0.0005 \mathrm{M}\), its pOH is: (a) \(8.2798\) (b) \(10.6990\) (c) \(12.7854\) (d) \(13.3344\)

Short Answer

Expert verified
The pOH is 10.6990 (option b).

Step by step solution

01

Determine pH from Hydrogen Ion Concentration

First, calculate the pH of the solution using the formula for pH: \[ \text{pH} = -\log[H^+] \]Substitute the given hydrogen ion concentration:\[ [H^+] = 0.0005 \text{ M} = 5 \times 10^{-4} \text{ M} \]Thus,\[ \text{pH} = -\log(5 \times 10^{-4}) = 3.3010 \]
02

Calculate pOH from pH

Next, use the relationship between pH and pOH to find pOH. The relationship is given by the equation:\[ \text{pH} + \text{pOH} = 14 \]Now, substitute the pH value into the equation:\[ 3.3010 + \text{pOH} = 14 \]\[ \text{pOH} = 14 - 3.3010 = 10.6990 \]

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Key Concepts

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

Hydrogen Ion Concentration
Understanding hydrogen ion concentration is the first step in calculating pH and ultimately pOH. The hydrogen ion concentration \([H^+]\) refers to the amount of hydrogen ions present in a solution. It is an indicator of a solution's acidity. In highly acidic solutions, you will find a higher concentration of hydrogen ions, whereas in basic (or alkaline) solutions, the concentration is lower.Hydrogen ion concentration is typically expressed in molarity (M), which is moles of hydrogen ions per liter of solution. For example, a concentration of \(0.0005 \, M\) means that there are \(0.0005\) moles of hydrogen ions in every liter of solution.
pH and pOH Relationship
The relationship between pH and pOH is a fundamental concept in chemistry, particularly in the study of acid-base balance. The pH scale measures the acidity or basicity of a solution. It typically ranges from 0 to 14, with lower values being more acidic, higher values being more basic, and 7 being neutral.The pOH, or "potential of hydroxide," complements the pH scale and is calculated using the formula: \[\text{pH} + \text{pOH} = 14 \] In essence, this formula tells us how the acidity (pH) and basicity (pOH) are interrelated. Suppose you know the pH of a solution (like \(3.3010\) in our exercise). In that case, you can easily find the pOH by rearranging the formula to: \[\text{pOH} = 14 - \text{pH}\]By understanding this, you realize that an increase in pH causes a decrease in pOH and vice versa, perfectly balancing the acidity and basicity of the solution.
Logarithmic Functions in Chemistry
Logarithms play a vital role in chemistry, especially when dealing with hydrogen ion concentrations and pH calculations. The pH of a solution is calculated using the formula: \[ \text{pH} = -\log[H^+] \]This equation involves the logarithm of the hydrogen ion concentration, which is why a small change in concentration results in a significant change in pH. Logarithms turn multiplicative relationships into additive ones, making them essential in dealing with the wide range of possible hydrogen ion concentrations.When calculating pH, it's helpful to transform \(0.0005 \, M\) into scientific notation: \(5 \times 10^{-4} \, M\). Applying the logarithmic rule to this scientific notation makes it easier to calculate the pH as shown in the exercise solution.
Acid-Base Balance
The concept of acid-base balance is crucial for maintaining the equilibrium in chemical reactions, biological systems, and industrial processes. Balancing acids and bases entails understanding how pH and pOH interrelate. Acids are substances that increase the hydrogen ion concentration of a solution, which decreases its pH. Bases, on the other hand, tend to increase the pOH by reducing the hydrogen ion concentration, thereby raising the pH. To maintain a stable system, it's important to understand how acids and bases neutralize each other. When you add an acid to a base or vice versa, their effects can either diminish or balance out depending on their concentrations, allowing for the sustainable management of pH levels. This balance is key in many chemical applications, from industrial processes to cellular metabolism.

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Most popular questions from this chapter

The ratio of \(\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}\) for the reaction: \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) is (a) 1 (b) RT (c) \((\mathrm{RT})^{1 / 2}\) (d) \((\mathrm{RT})^{-1 / 2}\)

At \(100^{\circ} \mathrm{C}\) the vapour density of nitrogen peroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) is \(26.8\). The percentage dissociation into \(\mathrm{NO}_{2}\) molecules is: (a) \(71.64 \%\) (b) \(61.57 \%\) (c) \(83.56 \%\) (d) \(67.39 \%\)

In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}:\) (a) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (c) \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)

At equilibrium total number of moles for the reaction \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\), if \(\alpha\) is degree of dissociation, are: (a) 2 (b) \(2-\alpha\) (c) 1 (d) \(\mathrm{I}-\alpha\)

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 \(K_{c}=4.9 \times 10^{-2}\). the value of \(K\) 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) \(4.9 \times 10^{-2}\)

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