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Calculate \(\left[\mathrm{OH}^{-}\right.\)] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{H}^{+}\right]=0.0505 \mathrm{M}\); (b) \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} \mathrm{M}\); (c) a solution in which \(\left[\mathrm{H}^{+}\right]\)is 1000 times greater than \([\mathrm{OH}]\).

Short Answer

Expert verified
a) [OH鈦籡 = \(1.98 \times 10^{-13}\) M, acidic b) [OH鈦籡 = \(4 \times 10^{-5}\) M, basic c) [OH鈦籡 = \(3.16 \times 10^{-6}\) M, acidic

Step by step solution

01

Use the ion product of water formula

Since we have [H鈦篯 = 0.0505 M, we can find [OH鈦籡 using the ion product of water formula, \(K_w = [H^+][OH^-]\), and solving for [OH鈦籡: \[[OH^-] = \frac{K_w}{[H^+]}\]
02

Insert the given values

Insert the given values of \(K_w=1 \times 10^{-14}\) and [H鈦篯 = 0.0505 M: \[[OH^-] = \frac{1 \times 10^{-14}}{0.0505}\]
03

Calculate [OH鈦籡 and identify the solution type

Calculate the value of [OH鈦籡 and identify if the solution is acidic, basic, or neutral: \[[OH^-] = 1.98 \times 10^{-13}\] Since [OH鈦籡 < [H鈦篯, the solution is acidic. #b. Calculate [OH鈦籡 for [H鈦篯 = 2.5 脳 10鈦宦光伆 M#
04

Use the ion product of water formula

Using the ion product of water formula, \(K_w = [H^+][OH^-]\), and solving for [OH鈦籡 with [H鈦篯 = 2.5 脳 10鈦宦光伆 M: \[[OH^-] = \frac{K_w}{[H^+]}\]
05

Insert the given values

Insert the given values of \(K_w=1 \times 10^{-14}\) and [H鈦篯 = 2.5 脳 10鈦宦光伆 M: \[[OH^-] = \frac{1 \times 10^{-14}}{2.5 \times 10^{-10}}\]
06

Calculate [OH鈦籡 and identify the solution type

Calculate the value of [OH鈦籡 and identify if the solution is acidic, basic, or neutral. \[ [OH^-] = 4 \times 10^{-5} \] Since [OH鈦籡 > [H鈦篯, the solution is basic. #c. Calculate [OH鈦籡 for [H鈦篯 being 1000 times greater than [OH鈦籡#
07

Set up the relationship between [H鈦篯 and [OH鈦籡

We are given that [H鈦篯 is 1000 times greater than [OH鈦籡. Let's represent this relationship: \[[H^+] = 1000[OH^-]\]
08

Use the ion product of water formula

Using the ion product of water formula, \(K_w = [H^+][OH^-]\), with [H鈦篯 = 1000 [OH鈦籡: \[K_w = 1000[OH^-]^2\]
09

Solve for [OH鈦籡

Solve for [OH鈦籡 by dividing both sides by 1000 and taking the square root: \[[OH^-] = \sqrt{\frac{K_w}{1000}}\]
10

Insert the given value of Kw and calculate [OH鈦籡

Insert the given value of \(K_w=1 \times 10^{-14}\) and calculate the value of [OH鈦籡: \[[OH^-] = \sqrt{\frac{1 \times 10^{-14}}{1000}} = 3.16 \times 10^{-6}\]
11

Identify the solution type

Since [H鈦篯 = 1000 [OH鈦籡, the solution is acidic.

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Key Concepts

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

pH Calculation
Understanding the pH scale is crucial for determining whether a solution is acidic, basic, or neutral. The pH is a logarithmic measure of the hydrogen ion concentration \( [H^+] \) in a solution. It is calculated using the formula \( pH = -\log[H^+] \). When \( [H^+] \) is known, as provided in our textbook exercise, you can easily compute the pH. For instance, if the hydrogen ion concentration is 0.0505 M, the pH is calculated as \( pH = -\log(0.0505) \), resulting in a pH less than 7, which indicates an acidic solution.

It's crucial for students to understand that a pH less than 7 is acidic, around 7 is neutral, and greater than 7 is basic. This corresponds inversely to the concentration of hydroxide ions \( [OH^-] \) in the solution, which can also be calculated using the relationship to the ion product of water. By mastering pH calculation, students can better comprehend the chemical properties of solutions and predict their behavior in various reactions.
Ion Product of Water
The ion product of water (\( K_w \) is a constant at a given temperature (25掳C value of \( 1 \times 10^{-14} \) that represents the product of the molar concentrations of hydrogen \( [H^+] \) and hydroxide \( [OH^-] \) ions in water. The formula is stated as \( K_w = [H^+][OH^-] \).

This constant value allows us to solve for either ion's concentration when the other is known. In the exercise, students are shown how to manipulate this expression to find the hydroxide ion concentration given the hydrogen ion concentration. For example, if \( [H^+] = 2.5 \times 10^{-10} M \) we solve for \( [OH^-] \) by rearranging the equation to \( [OH^-] = \frac{K_w}{[H^+]} \), inserting the known values, and solving for \( [OH^-] \). This is a fundamental concept in acid-base chemistry and is essential for students to be able to shift between \( [H^+] \) and \( [OH^-] \) calculations.
Acidic and Basic Solutions
The nature of a solution, whether it is acidic or basic, is determined by its pH, which is in turn dependent on the relative concentrations of hydrogen \( [H^+] \) and hydroxide \( [OH^-] \) ions.

An acidic solution has a higher concentration of hydrogen ions than hydroxide ions \( [H^+] > [OH^-] \), while a basic solution has a lower concentration of hydrogen ions than hydroxide ions \( [H^+] < [OH^-] \). When the concentrations of \( [H^+] \) and \( [OH^-] \) are equal, the solution is neutral. In our textbook problem, the nature of the solution is deduced after calculating the \( [OH^-] \) concentration. For instance, in part a, after finding \( [OH^-] = 1.98 \times 10^{-13} \), since this value is less than the given \( [H^+] \), the solution is acidic. Conversely, in part b, where \( [OH^-] > [H^+] \) the solution is basic.

It is important for students to connect the pH and \( [OH^-] \)-concentration calculations with the concept of acidity and basicity to develop a holistic understanding of chemical solutions.

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Most popular questions from this chapter

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{p} K_{a}\) of butyric acid is \(4.84\). (a) Calculate the \(\mathrm{p} K_{b}\) for the butyrate ion. (b) Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.050 \mathrm{M}\) solution of sodium butyrate.

Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{\text {; }}\) (b) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;\) (c) formate ion, \(\mathrm{CHO}_{2}^{-}\).

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid ( \(\left.\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S}\) because Te is more electronegative than \(\mathrm{S}\).

(a) Which of the following is the stronger Br酶nsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2}\) ? (b) Which is the stronger BronstedLowry base, \(\mathrm{HS}^{-}\)or \(\mathrm{HSO}_{4}^{-}\)?

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