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Chapter 16: Problem 29

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 M ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;\) (c) a solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\) .

Short Answer

Expert verified
For the given solutions: (a) \([\mathrm{H}^{+}] = 2.22 \times 10^{-12}\,\text{M}\) and the solution is basic; (b) \([\mathrm{H}^{+}] = 1.14 \times 10^{-6}\,\text{M}\) and the solution is acidic; (c) \([\mathrm{H}^{+}] = 1.0 \times 10^{-7}\,\text{M}\) and the solution is neutral.

Step by step solution

01

Find \([\mathrm{H}^{+}]\)

Given that \([\mathrm{OH}^{-}] = 0.00045\,\text{M}\), we can find \([\mathrm{H}^{+}]\) using the K_w expression: \[\left[\mathrm{H}^{+}\right]=\frac{K_\text{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0\times10^{-14}}{0.00045}\]
02

Calculate \([\mathrm{H}^{+}]\) and determine the solution type

Calculating the concentration of hydrogen ions: \[\left[\mathrm{H}^{+}\right] = 2.22 \times 10^{-12}\,\text{M}\] Since \([\mathrm{H}^{+}] < [\mathrm{OH}^{-}]\), the solution is basic. #Case (b):#
03

Find \([\mathrm{H}^{+}]\)

Given that \([\mathrm{OH}^{-}]=8.8\times10^{-9}\,\text{M}\), we can find \([\mathrm{H}^{+}]\) using the K_w expression: \[\left[\mathrm{H}^{+}\right]=\frac{K_\text{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0\times10^{-14}}{8.8\times10^{-9}}\]
04

Calculate \([\mathrm{H}^{+}]\) and determine the solution type

Calculating the concentration of hydrogen ions: \[\left[\mathrm{H}^{+}\right] = 1.14 \times 10^{-6}\,\text{M}\] Since \([\mathrm{H}^{+}] > [\mathrm{OH}^{-}]\), the solution is acidic. #Case (c):#
05

Write the given relation

We are given that \([\mathrm{OH}^{-}]\) is 100 times greater than \([\mathrm{H}^{+}]\): \[\left[\mathrm{OH}^{-}\right] = 100\left[\mathrm{H}^{+}\right]\]
06

Find \([\mathrm{H}^{+}]\) using the K_w expression

Using the K_w expression and substituting the given relation: \[\left[\mathrm{H}^{+}\right]\left(100\left[\mathrm{H}^{+}\right]\right) = 1.0 \times 10^{-14}\]
07

Solve for \([\mathrm{H}^{+}]\) and determine the solution type

Solving the equation for \([\mathrm{H}^{+}]\) we get: \[\left[\mathrm{H}^{+}\right] = 1.0 \times 10^{-7}\,\text{M}\] Since \([\mathrm{H}^{+}] = [\mathrm{OH}^{-}]\) (given their ratio is 100), the solution is neutral.

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Key Concepts

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

Understanding Acidic Solutions
An acidic solution is one in which the concentration of hydrogen ions \([\mathrm{H}^{+}]\) is greater than the concentration of hydroxide ions \([\mathrm{OH}^{-}]\).
This means that the solution has a pH less than 7. pH is a measure of how acidic or basic a solution is, with lower values indicating higher acidity.
Acidic solutions can be found in various places, like vinegar, orange juice, and even in our stomachs.
  • An excess of \([\mathrm{H}^{+}]\) ions in a solution means more acidity.
  • The formula for calculating \([\mathrm{H}^{+}]\) is helpful to determine acidity: \([\mathrm{H}^{+}] = \frac{{K_{\text{w}}}}{{[\mathrm{OH}^{-}]}}\).
  • In calculations, if \([\mathrm{H}^{+}] > [\mathrm{OH}^{-}]\), the solution is acidic.
Simply put, knowing the balance between \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) helps identify and understand the nature of acidic solutions.
Exploring Basic Solutions
A basic solution, often referred to as alkaline, has a higher concentration of hydroxide ions \([\mathrm{OH}^{-}]\) compared to hydrogen ions \([\mathrm{H}^{+}]\).
This results in a pH greater than 7. Such solutions are common in cleaning products like soaps and detergents.
Bases can neutralize acids, hence they're used in various applications to balance pH.
  • The formula for \([\mathrm{H}^{+}]\) helps in identifying basic solutions: \([\mathrm{H}^{+}] = \frac{{K_{\text{w}}}}{{[\mathrm{OH}^{-}]}}\).
  • If \([\mathrm{H}^{+}] < [\mathrm{OH}^{-}]\), the solution is basic.
  • Basic solutions feel slippery and can change red litmus paper blue.
Understanding these properties helps us use basic solutions effectively and safely in our everyday life.
Defining Neutral Solutions
Neutral solutions are characterized by an equal concentration of hydrogen ions \([\mathrm{H}^{+}]\) and hydroxide ions \([\mathrm{OH}^{-}]\), resulting in a pH of exactly 7.
The most common example of a neutral solution is pure water.
These solutions are neither acidic nor basic, making them perfect benchmarks in scientific studies.
  • Using the equation \([\mathrm{H}^{+}] = [\mathrm{OH}^{-}]\), we identify a solution as neutral.
  • This balance is crucial in many natural and industrial processes for maintaining stability.
  • Neutral solutions do not alter the color of litmus paper, staying unaltered like water.
The balance in neutral solutions provides a foundation for understanding both acidic and basic solutions.

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Most popular questions from this chapter

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a})\mathrm{CH}_{3} \mathrm{COO}^{-},(\mathbf{b}) \mathrm{HCO}_{3}^{-},(\mathbf{c}) \mathrm{O}^{2-},(\mathbf{d}) \mathrm{Cl}^{-},(\mathbf{e}) \mathrm{NH}_{3}\)

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Deuterium oxide (\(\mathrm{D}_{2} \mathrm{O},\) where \(\mathrm{D}\) is deuterium, the hydrogen-2 isotope) has an ion-product constant, \(K_{w}\) , of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure (neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q) ?\) What is its conjugate base?

The iodate ion is reduced by sulfite according to the following reaction: $$\mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}^{2-}(a q)$$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}^{-}\) , first order in \(\mathrm{SO}_{3}^{2-}\) , and first order in \(\mathrm{H}^{+}\) . (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the pH is lowered from 5.00 to 3.50\(?\) Does the reaction proceed more quickly or more slowly at the lower pH? (c) By using the concepts discussed in Section 14.6, ex-plain how the reaction can be pH-dependent even though H' does not appear in the overall reaction.
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