# 8.27: Brønsted-Lowry Acids and Bases: Calculating Acid/Base Concentrations of Non-Neutral Solutions

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- 233580

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

As will be discussed in the remaining sections of this chapter, chemists have derived two equations that can be used to calculate the amounts of acid and base that are present in a solution. Chemists initially studied a "solution" of "pure" liquid water, H_{2}O, which was classified as "neutral." Because no acidic or basic solutes were dissolved in this "solution," scientists expected to detect only a single chemical, water, H_{2}O. Instead, these researchers discovered that very small amounts of acid *and* base were present in *all* of the water samples that were tested. This unexpected result can be explained by predicting the products that are generated during the autoionization of water, in which a proton, H^{+1}, is spontaneously transferred from one water molecule to another. The Brønsted-Lowry acid/base equation that symbolically-represents this reaction was developed in Section 8.25 and is reproduced below.

\(\ce{H_2O}\) \(\left( l \right)\) + \(\ce{H_2O}\) \(\left( l \right)\) \(\longrightleftharpoons\) \(\ce{OH^{–1}}\) \(\left( aq \right)\) + \(\ce{H_3O^{+1}}\) \(\left( aq \right)\)

Because the conjugate products that are generated during this reaction are found in *all *aqueous solutions that contain basic or acidic solutes, chemists chose to quantify the amounts of base and acid in a solution by measuring the amounts of hydroxide ions, OH^{–1}, and hydronium ions, H_{3}O^{+1}, respectively, that are present in that solution. Since the Brønsted-Lowry solutes that are being investigated are, by definition, contained in *solutions*, the measured amounts of these ions must be reported as concentrations. In contrast to mass percent, volume percent, and mass/volume percent concentration calculations, which are typically simplified and presented as "end-result" values, molarities are molar quantities and, therefore, can be related to the other molar standards that have been presented in this textbook. Recall that, in order to indicate the identity of a solute in concentration-based calculations, the "Molarity" variable can be replaced by the chemical formula of that solute, enclosed in square brackets. Therefore, the molarities of hydroxide ions, OH^{–1}, and hydronium ions, H_{3}O^{+1}, that are present in a solution can be represented as [OH^{–1}] and [H_{3}O^{+1}], respectively.

An equilibrium arrow is incorporated into the acid/base equation that is shown above, because the reactants are classified as a weak acid and a weak base, respectively. Recall that weak acids and weak bases only partially ionize in solution and, subsequently, undergo both "forward" dissociation and "reverse" recombination processes when solvated. While some solutes favor dissociation, and therefore, ionize to generate relatively-high amounts of products, other chemicals, which favor recombination, produce solutions that contain a greater proportion of reactants. Therefore, in order to determine the concentrations of reactants and products that are present in a solution, an **equilibrium equation** that corresponds to the given chemical reaction must be derived. In order to generate this type of equation, the molar concentrations of all *aqueous* and *gaseous products* must be multiplied. In the Brønsted-Lowry acid/base equation that is shown above, both of the products are generated in the aqueous state of matter and, therefore, the molarities of the hydroxide ion, OH^{–1}, and the hydronium ion, H_{3}O^{+1}, must be incorporated into the equilibrium equation for this reaction, as shown below.

\(\rm{[OH^{–1}]}\) × \(\rm{[H_3O^{+1}]}\)

These quantities must then be divided by the molar concentrations of all *aqueous* and *gaseous reactants*. In the reaction equation that is shown above, water, H_{2}O, reacts a liquid and, consequently, is not represented in the equilibrium equation that is being developed.

Finally, the resultant mathematical statement must be equated to an **equilibrium constant**,** ****K**, which indicates the relative concentrations of the reactants and products that are present in a solution. Because the reaction that is shown above represents the autoionization of *water*, H_{2}O, a "w" is written as a subscript on this equilibrium constant, as shown below.

\(\rm{K_{w}}\) = \(\rm{[OH^{–1}]}\) × \(\rm{[H_3O^{+1}]}\)

This equation is known as the **Ion-Product Constant Equation**, because it was derived by multiplying the concentrations of two charged particles, or *ions*, and equating the resultant *mathematical product* to a *constant*. The variables in this equation can be written in any order, and the multiplication symbol can be omitted, as shown below.

\(\rm{K_{w}}\) = \(\rm{[H_3O^{+1}]}\)\(\rm{[OH^{–1}]}\)

The value of K_{w} can be determined by replacing the [H_{3}O^{+1}] and [OH^{–1}] variables in this equation with known numerical values. As stated in the previous section of this chapter, the experimentally-determined molarity of hydronium ions, H_{3}O^{+1}, is 1.00 x 10^{–7} *M* in a neutral solution. Therefore, because the molarity of hydroxide ions, OH^{–1}, must be equal to the molarity of hydronium ions, H_{3}O^{+1}, in a neutral solution,

\(\rm{K_{w}}\) = (\({1.00 \times 10^{–7} M}\))(\({1.00 \times 10^{–7} M}\))

\(\rm{K_{w}}\) = \({1.00 \times 10^{–14}}\)

When using a calculator to determine K_{w}, 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 section of this chapter, the value of K_{w}, 1.00 x 10^{–14}, which is expressed in scientific notation, must be reported to three significant figures. Finally, while the unit that should result upon multiplying two molar concentrations should be "molarity squared," or "*M ^{2}*," equilibrium constants are typically expressed as unitless quantities. Therefore, the concentration units are eliminated during the simplification of the Ion-Product Constant Equation,

*.*

*even though these units do not cancel, mathematically*While the value of K_{w} was initially determined by multiplying the molarities of hydronium ions, H_{3}O^{+1}, and hydroxide ions, OH^{–1}, that are present in a neutral solution, equilibrium constants are universal constants and, therefore, can be used to determine the concentrations of reactants and products that are present in *any* solution. As a result, the Ion-Product Constant Equation is commonly written by replacing the "K_{w}" variable with its calculated numerical value, 1.00 x 10^{–14}, as shown below.

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

Finally, the molarities of hydroxide ions, OH^{–1}, and hydronium ions, H_{3}O^{+1}, that are present in a solution can be compared to determine whether the amount of acid or the quantity of base predominates in and, therefore, dictates the characteristics of, that solution. As stated previously, the molarity of hydronium ions, H_{3}O^{+1}, and the molarity of hydroxide ions, OH^{–1}, must be equal in a neutral solution. In order to be classified as acidic, a homogeneous mixture must contain more acidic hydronium ions, H_{3}O^{+1}, than basic hydroxide ions, OH^{–1}. In contrast, a basic solution will have a higher concentration of hydroxide ions, OH^{–1}. The values that are shown above are written in scientific notation, which, as stated previously, is used to represent very large or very small numerical quantities without the unnecessary placeholder zeroes that must be present when those numbers are written in decimal format. Since the value of the power corresponds to the number of decimal places that are eliminated, the magnitude of a number that is written in scientific notation *increases *as the value of a *negative power* decreases.

For example, classify a solution in which hydronium ions and hydroxide ions are present in 1.00 x 10^{–8} *M* and 1.00 x 10^{–6} *M* concentrations, respectively, as acidic, basic, or neutral.

As stated above, the value of the power that is used to write a number in scientific notation corresponds to the number of decimal places that are eliminated by using this type of representation. Therefore, the magnitude of a number that is written in scientific notation *increases *as the value of a *negative power* decreases. Since –6 is a smaller negative exponent than –8, the given molarity of hydroxide ions, OH^{–1}, 1.00 x 10^{–6} *M*, which can also be written as 0.00000100 *M*, is greater than the concentration of hydronium ions, H_{3}O^{+1}, 1.00 x 10^{–8} *M*, which can also be represented as 0.0000000100 *M*. Therefore, the solution is classified as basic.

Calculate the concentration of hydronium ions that is present in a solution that has a 2.63 x 10^{–13} *M* concentration of hydroxide ions, and classify the solution as acidic, basic, or neutral.

**Solution**

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. Therefore, in order to calculate the concentration of hydronium ions, H_{3}O^{+1}, that is present in a solution, the molar concentration of hydroxide ions, OH^{–1}, 2.63 x 10^{–13} *M*, is incorporated into this 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 section 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*.

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

\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)(\({2.63 \times 10^{–13} M}\))

\(\rm{[H_3O^{+1}]}\) = \({3.80228... \times 10^{–2} M} ≈ {3.80 \times 10^{–2} M}\)

As stated above, the magnitude of a number that is written in scientific notation *increases *as the value of a *negative power* decreases. Since –2 is a smaller negative exponent than –13, the calculated concentration of hydronium ions, H_{3}O^{+1}, 3.80 x 10^{–2} *M*, which can also be represented as 0.0380 *M*, is greater than the given molarity of hydroxide ions, OH^{–1}, 2.63 x 10^{–13} *M*, which can also be written as 0.000000000000263 *M*. Therefore, the solution is classified as acidic.