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A certain solution has \(\mathrm{pH} 3.89\) at \(0^{\circ} \mathrm{C}\). Find \(\mathrm{pOH}\) and \(\left[\mathrm{OH}^{-}\right]\).

Short Answer

Expert verified
The pOH of the solution is 10.11 and the concentration of hydroxide ions, [OH⁻], is approximately \(7.94 \times 10^{-11}\) M.

Step by step solution

01

Find pOH using the given pH value

The given pH of the solution is 3.89. To find the pOH, we can use the equation pH + pOH = 14: \(3.89 + \textrm{pOH} = 14\) Now, solve for pOH: \(\textrm{pOH} = 14 - 3.89\) \(\textrm{pOH} = 10.11\)
02

Find [OH⁻] using the pOH value

Now that we have found the pOH value, we can use the equation relating pOH and [OH⁻] to find the concentration of hydroxide ions: \(\textrm{pOH} = -\textrm{log}_{10} [\textrm{OH}^{-}]\) Substitute the pOH value obtained in step 1: \(10.11 = -\textrm{log}_{10} [\textrm{OH}^{-}]\) To find [OH⁻], we need to rearrange the equation and solve for [OH⁻]: \([\textrm{OH}^{-}] = 10^{-\textrm{pOH}}\) Substitute the pOH value: \([\textrm{OH}^{-}] = 10^{-10.11}\) Now, calculate the concentration of hydroxide ions: \([\textrm{OH}^{-}] \approx 7.94 \times 10^{-11} \, \textrm{M}\) So, the pOH of the solution is 10.11 and the concentration of hydroxide ions, [OH⁻], is approximately 7.94 x 10⁻¹¹ M.

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Key Concepts

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

pH Scale
The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with 7 being neutral, below 7 acidic, and above 7 basic. The letters 'pH' stand for 'potential of hydrogen' and refer to the concentration of hydrogen ions, \( H^+ \), in a solution.

When talking about the pH scale, it's important to understand that each whole pH value below 7 is ten times more acidic than the next higher value. For example, a solution with a pH of 3 is ten times more acidic than a solution with a pH of 4. Conversely, a solution with a pH of 11 is ten times more basic, or alkaline, than one with a pH of 10. This logarithmic nature means the pH scale compresses a wide range of ion concentrations into a manageable scale.
pOH
pOH is a measure, like pH, but it represents the concentration of hydroxide ions, \( OH^- \) , in a solution. Just like pH, pOH is also a logarithmic scale and is calculated using the negative logarithm of the hydroxide ion concentration. The relationship between pH and pOH is neatly summarized by the equation \( pH + pOH = 14 \), which is true at 25° C, in pure water.

Understanding pOH is critical, especially when dealing with solutions that are basic or alkaline. While pH is more commonly referred to in day-to-day contexts, pOH is equally important in academic and professional scenarios to precisely articulate the nature of a basic solution.
Hydroxide Ion Concentration
Hydroxide ion concentration \( [OH^-] \) determines the basicity of a solution. It is the amount of hydroxide ions, usually measured in moles per liter (M), in an aqueous solution. Higher concentrations indicate a more basic solution, and lower concentrations indicate a less basic, or more acidic, solution.

To calculate the concentration of hydroxide ions from pOH, you use the formula \( [OH^-] = 10^{-pOH} \). Just like pH, the relationship between pOH and the concentration of hydroxide ions is inverse and logarithmic. This means if pOH decreases, the hydroxide ion concentration increases, and vice versa.
Acid-Base Equilibrium
Acid-base equilibrium pertains to the balance between acids (substances that donate protons \( H^+ \)) and bases (substances that accept protons or donate hydroxide ions \( OH^- \)) in a solution. This concept is fundamental in understanding the behavior of substances in water, which either tend to release hydrogen ions, becoming acidic, or absorb them, becoming basic.

The equilibrium is quantitatively characterized by the dissociation constants \( K_a \) for acids and \( K_b \) for bases, which are specific to each substance. The product of these constants for a conjugate acid-base pair is always equal to the ion-product constant for water \( K_w \), demonstrating the intrinsic relationship between acids, bases, and the medium they are in.

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Most popular questions from this chapter

A common method for commercially peeling potatoes is to soak them for \(1-5\) minutes in a \(10-20 \%\) solution of \(\mathrm{NaOH}\) (molecular weight \(=40.0 \mathrm{~g} /\) mole \()\) at \(60-88^{\circ} \mathrm{C}\), and to spray off the peel once the potatoes are removed from solution. As an economy measure, a manufacturer titrates the \(\mathrm{NaOH}\) solution with standardized \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (molecular weight \(=98.0\) \(\mathrm{g} /\) mole ) at the end of each day to determine whether the solution is still capable of peeling potatoes. If, at the end of one day, he finds that it takes \(64.0 \mathrm{ml}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to titrate a \(10.0 \mathrm{ml}\) sample of \(\mathrm{NaOH}\) solution to neutrality, what concentration of \(\mathrm{NaOH}\) did he find?

\(50 \mathrm{ml}\) of rhubarb juice is titrated against \(0.25 \mathrm{~N} \mathrm{NaOH} .20 \mathrm{ml}\) of \(\mathrm{NaOH}\) solution is required for neutralization. Assuming the acidity of the juice is due to oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) determine (a) the weight of oxalic acid per liter of juice, (b) normality of the juice.

A saturated solution of the strong electrolyte \(\mathrm{Ca}(\mathrm{OH})_{2}\) is prepared by adding sufficient water to \(5.0 \times 10^{-4}\) mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) to form \(100 \mathrm{ml}\) of solution. What is the \(\mathrm{pH}\) of this solution?

\(0.1\) moles of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) are added to \(1.0\) moles of \(\mathrm{NH}_{3}\) and then diluted to \(1000 \mathrm{ml}\). Find the concentration of \(\mathrm{Cu}^{2+}\) formed. $$ \mathrm{K}_{\mathrm{eq}}=8.5 \times 10^{-13} $$

Determine the \(\mathrm{pH}\) of each of the following solutions: (a) \(0.20 \mathrm{M} \mathrm{HCl}\), (b) \(0.10 \mathrm{M} \mathrm{NaOH}\).

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