# 8.30: Brønsted-Lowry Acids and Bases: Additional Acid/Base Concentration and pH Examples

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- 233678

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Apply the Ion-Product Constant Equation to calculate the concentrations of hydronium ions or hydroxide ions that are present in a solution.
- Apply the pH Equation to calculate the pH of a solution.
- Apply the anti-log pH Equation to calculate the concentration of hydronium ions that are present in a solution.
- Classify a solution as acidic, basic, or neutral by comparing the concentrations of hydronium ions and hydroxide ions that are present in that solution.
- Classify a solution as acidic, basic, or neutral by comparing the pH of that solution to the pH scale.

Calculate the concentration of hydronium ions that are present in, as well as the pH of, a solution that has a 8.13 x 10^{–5} *M* concentration of hydroxide ions, and classify the solution as acidic, basic, or neutral.

**Answer**- The molarities of hydronium ions, H
_{3}O^{+1}, and hydroxide ions, OH^{–1}, that are present in a solution are multiplicatively-related in the Ion-Product Constant Equation. Furthermore, the molarity of hydronium ions, H_{3}O^{+1}, that is present in a solution is related to the pH of that solution by the pH Equation. Finally, the molarity of hydroxide ions, OH^{–1}, that is present in a solution cannot be directly related to the pH of that solution. Therefore, the given molarity of hydroxide ions, OH^{–1}, must first be used to calculate the concentration of hydronium ions, H_{3}O^{+1}, that are present in the solution, and the resultant value can be used to determine the pH of the solution.In order to calculate the concentration of hydronium ions, H

_{3}O^{+1}, that are present in this solution, the molar concentration of hydroxide ions, OH^{–1}, 8.13 x 10^{–5}*M*, is incorporated into the Ion-Product Constant Equation, as shown below. The numerical solution is calculated by dividing the value of K_{w}, 1.00 x 10^{–14}, by the given molarity, and, when using a calculator, any quantity that is expressed in scientific notation should be offset by parentheses. In order to align with the standards that were established in the previous sections of this chapter, the final answer, which is expressed in scientific notation, is reported to three significant figures. Finally, the numerical solution corresponds to a molarity, and, therefore, is labeled with a unit of*"**M*,*"*.*even though this unit is not established through the mathematical process that is described above*\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)\(\rm{[OH^{–1}]}\)

\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)(\({8.13 \times 10^{–5} M}\))

\(\rm{[H_3O^{+1}]}\) = \({1.23001... \times 10^{–10} M} ≈ {1.23 \times 10^{–10} M}\)In order to calculate the pH of this solution, the calculated molar concentration of hydronium ions, H

_{3}O^{+1}, 1.23 x 10^{–10}*M*, is incorporated into the pH Equation, as shown below. When using a calculator, the negative key, (–), must be used, the subscript value of "10" should not be entered, and any quantity that is expressed in scientific notation should be offset by parentheses. Furthermore, in order to align with the standards that were established in the previous sections of this chapter, a pH value, which is, by mathematical definition, a unitless quantity, is recorded to the hundredths place.\(\rm{pH}\) = \({–\rm{log_{10}}}\)\(\rm{[H_3O^{+1}]}\)

Finally, in order to classify this solution as acidic, basic, or neutral, the hydronium, H

\(\rm{pH}\) = \({–\rm{log_{10}}}\)(\({1.23 \times 10^{–10} M}\))

\(\rm{pH}\) = \({9.91009...} ≈ {9.91}\)_{3}O^{+1}, and hydroxide, OH^{–1}, ion concentrations can be compared to one another. Alternatively, the calculated pH can be analyzed, relative to the values on the pH Scale.The magnitude of a number that is written in scientific notation

*increases*as the value of a*negative power*decreases. Since –5 is a smaller negative exponent than –10, the given molarity of hydroxide ions, OH^{–1}, 8.13 x 10^{–5}*M*, which can also be written as 0.0000813*M*, is greater than the calculated concentration of hydronium ions, H_{3}O^{+1}, 1.23 x 10^{–10}*M*, which can also be represented as 0.000000000123*M*. Therefore, the solution is classified as basic.

The pH value of a neutral solution is*exactly*7.00, any solution that has a pH that is*greater than*7.00 can be classified as basic, and an acidic solution must have a pH value that is*less than than*7.00. Therefore, because the calculated pH, 9.91, is*greater than*7.00, this solution is classified as basic.

Calculate the concentrations of hydronium ions and hydroxide ions that are present in a solution that has a reported pH value of 13.54, and classify the solution as acidic, basic, or neutral.

**Answer**- The molarities of hydronium ions, H
_{3}O^{+1}, and hydroxide ions, OH^{–1}, that are present in a solution are multiplicatively-related in the Ion-Product Constant Equation. Furthermore, the molarity of hydronium ions, H_{3}O^{+1}, that is present in a solution is related to the pH of that solution by the pH Equation. Finally, the molarity of hydroxide ions, OH^{–1}, that is present in a solution cannot be directly related to the pH of that solution. Therefore, the given pH must first be used to calculate the concentration of hydronium ions, H_{3}O^{+1}, that are present in the solution, and the resultant value can be used to determine the molarity of hydroxide ions, OH^{–1}, that are present in the solution.In order to calculate the molar concentration of hydronium ions, H

_{3}O^{+1}, from a pH, the given value, 13.54, is incorporated into the anti-log pH Equation, as shown below. As established in the previous sections of this chapter, the numerical solution, which corresponds to a concentration of hydronium ions, H_{3}O^{+1}, must be expressed in scientific notation and reported to three significant figures. Therefore, in order to calculate a properly-formatted concentration value, the quantity on the right side of the second equation that is shown below must be entered into a calculator. Finally, the resultant value is a molarity, and, therefore, is labeled with a unit of*"**M*,*"**even though this unit is not established through the mathematical processes that are described*.*above*\(\rm{[H_3O^{+1}]}\) = \({10^{\rm{–pH}}}\)

\(\rm{[H_3O^{+1}]}\) = \({10^{–13.54}}\)

\(\rm{[H_3O^{+1}]}\) = \({2.88403... \times 10^{–14} M} ≈ {2.88 \times 10^{–14} M}\)In order to calculate the concentration of hydroxide ions, OH

^{–1}, that are present in this solution, the calculated molar concentration of hydronium ions, H_{3}O^{+1}, 2.88 x 10^{–14}*M*, is incorporated into the Ion-Product Constant Equation, as shown below. The numerical solution is calculated by dividing the value of K_{w}, 1.00 x 10^{–14}, by the calculated molarity, and, when using a calculator, any quantity that is expressed in scientific notation should be offset by parentheses. In order to align with the standards that were established in the previous sections of this chapter, the final answer, which is expressed in scientific notation, is reported to three significant figures. Finally, the numerical solution corresponds to a molarity, and, therefore, is labeled with a unit of*"**M*,*"*.*even though this unit is not established through the mathematical process that is described above*\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)\(\rm{[OH^{–1}]}\)

Finally, in order to classify this solution as acidic, basic, or neutral, the hydronium, H

\({1.00 \times 10^{–14}}\) = (\({2.88 \times 10^{–14} M}\))\(\rm{[OH^{–1}]}\)

\(\rm{[OH^{–1}]}\) = \({3.47222... \times 10^{–1} M} ≈ {3.47 \times 10^{–1} M}\)_{3}O^{+1}, and hydroxide, OH^{–1}, ion concentrations can be compared to one another. Alternatively, the given pH can be analyzed, relative to the values on the pH Scale.The magnitude of a number that is written in scientific notation

*increases*as the value of a*negative power*decreases. Since –1 is a smaller negative exponent than –14, the calculated molarity of hydroxide ions, OH^{–1}, 3.47 x 10^{–1}*M*, which can also be written as 0.347*M*, is greater than the calculated concentration of hydronium ions, H_{3}O^{+1}, 2.88 x 10^{–14}*M*, which can also be represented as 0.0000000000000288*M*. Therefore, the solution is classified as basic.

The pH value of a neutral solution is*exactly*7.00, any solution that has a pH that is*greater than*7.00 can be classified as basic, and an acidic solution must have a pH value that is*less than than*7.00. Therefore, because the given pH, 13.54, is*greater than*7.00, this solution is classified as basic.

Calculate the concentration of hydroxide ions that are present in, as well as the pH of, a solution that has a 1.07 x 10^{–7} *M* concentration of hydronium ions, and classify the solution as acidic, basic, or neutral.

**Answer**- The molarities of hydronium ions, H
_{3}O^{+1}, and hydroxide ions, OH^{–1}, that are present in a solution are multiplicatively-related in the Ion-Product Constant Equation. Furthermore, the molarity of hydronium ions, H_{3}O^{+1}, that is present in a solution is related to the pH of that solution by the pH Equation. Finally, the molarity of hydroxide ions, OH^{–1}, that is present in a solution cannot be directly related to the pH of that solution. Therefore, the given molarity of hydronium ions, H_{3}O^{+1}, can be directly applied to calculate both the concentration of hydroxide ions, OH^{–1}, that are present in the solution*and*the pH of the solution.In order to calculate the concentration of hydroxide ions, OH

^{–1}, that are present in this solution, the molar concentration of hydronium ions, H_{3}O^{+1}, 1.07 x 10^{–7}*M*, is incorporated into the Ion-Product Constant Equation, as shown below. The numerical solution is calculated by dividing the value of K_{w}, 1.00 x 10^{–14}, by the given molarity, and, when using a calculator, any quantity that is expressed in scientific notation should be offset by parentheses. In order to align with the standards that were established in the previous sections of this chapter, the final answer, which is expressed in scientific notation, is reported to three significant figures. Finally, the numerical solution corresponds to a molarity, and, therefore, is labeled with a unit of*"**M*,*"*.*even though this unit is not established through the mathematical process that is described above*\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)\(\rm{[OH^{–1}]}\)

\({1.00 \times 10^{–14}}\) = (\({1.07 \times 10^{–7} M}\))\(\rm{[OH^{–1}]}\)

\(\rm{[OH^{–1}]}\) = \({9.34579... \times 10^{–8} M} ≈ {9.35 \times 10^{–8} M}\)In order to calculate the pH of this solution, the molar concentration of hydronium ions, H

_{3}O^{+1}, 1.07 x 10^{–7}*M*, is incorporated into the pH Equation, as shown below. When using a calculator, the negative key, (–), must be used, the subscript value of "10" should not be entered, and any quantity that is expressed in scientific notation should be offset by parentheses. Furthermore, in order to align with the standards that were established in the previous sections of this chapter, a pH value, which is, by mathematical definition, a unitless quantity, is recorded to the hundredths place.\(\rm{pH}\) = \({–\rm{log_{10}}}\)\(\rm{[H_3O^{+1}]}\)

Finally, in order to classify this solution as acidic, basic, or neutral, the hydronium, H

\(\rm{pH}\) = \({–\rm{log_{10}}}\)(\({1.07 \times 10^{–7} M}\))

\(\rm{pH}\) = \({6.970616...} ≈ {6.97}\)_{3}O^{+1}, and hydroxide, OH^{–1}, ion concentrations can be compared to one another. Alternatively, the calculated pH can be analyzed, relative to the values on the pH Scale.The magnitude of a number that is written in scientific notation

*increases*as the value of a*negative power*decreases. Since –7 is a smaller negative exponent than –8, the given molarity of hydronium ions, H_{3}O^{+1}, 1.07 x 10^{–7}*M*, which can also be written as 0.000000107*M*, is greater than the calculated concentration of hydroxide ions, OH^{–1}, 9.35 x 10^{–8}*M*, which can also be represented as 0.0000000935*M*. Therefore, the solution is classified as acidic.

The pH value of a neutral solution is*exactly*7.00, any solution that has a pH that is*greater than*7.00 can be classified as basic, and an acidic solution must have a pH value that is*less than than*7.00. Therefore, because the calculated pH, 6.97, is*less than*7.00, this solution is classified as acidic.