Polyprotic Acids

Learning Outcomes

Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton

We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO₃, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

[latex]\begin{array}{c}\text{HCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\\ {\text{HNO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NO}}_{3}{}^{-}\left(aq\right)\\ \text{HCN}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\leftrightharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\end{array}[/latex]

Even though it contains four hydrogen atoms, acetic acid, CH₃CO₂H, is also monoprotic because only the hydrogen atom from the carboxyl group (COOH) reacts with bases:

Similarly, monoprotic bases are bases that will accept a single proton.

Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:

[latex]\begin{array}{lll} \text{First ionization: }{\text{H}}_{2}{\text{SO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HSO}}_{4}{}^{-}\left(aq\right)&{}&{K}_{\text{a}1}=\text{more than }{10}^{2+}\text{; complete dissociation}\\ \text{Second ionization: }{\text{HSO}}_{4}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)&{}&{K}_{\text{a}2}=1.2\times {10}^{-2}\end{array}[/latex]

This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, H₂CO₃, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

[latex]\text{First ionization: }{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,\,{K}_{{\text{H}}_{2}{\text{CO}}_{3}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{HCO}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=4.3\times {10}^{-7}[/latex]

The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.

[latex]\text{Second ionization: }{\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\,\,\,\,\,\,\,\,\,\,\,\,{K}_{{\text{HCO}}_{3}{}^{-}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}=4.7\times {10}^{-11}[/latex]

[latex]{K}_{{\text{H}}_{2}{\text{CO}}_{3}}[/latex] is larger than [latex]{K}_{{\text{HCO}}_{3}{}^{-}}[/latex] by a factor of 10⁴, so H₂CO₃ is the dominant producer of hydronium ion in the solution. This means that little of the [latex]{\text{HCO}}_{3}{}^{-}[/latex] formed by the ionization of H₂CO₃ ionizes to give hydronium ions (and carbonate ions), and the concentrations of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and [latex]{\text{HCO}}_{3}{}^{-}[/latex] are practically equal in a pure aqueous solution of H₂CO₃.

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.

Example 1: Ionization of a Diprotic Acid

When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO₂ reacts with water to form carbonic acid, H₂CO₃. What are [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex], [latex]\left[{\text{HCO}}_{3}{}^{-}\right][/latex], and [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] in a saturated solution of CO₂ with an initial [H₂CO₃] = 0.033 M?

[latex]{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{-}\left(aq\right)\qquad{K}_{\text{a}1}=4.3\times {10}^{-7}[/latex]
[latex]{\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\qquad{K}_{\text{a2}}=4.7\times {10}^{-11}[/latex]

Show Solution

As indicated by the ionization constants, H₂CO₃ is a much stronger acid than [latex]{\text{HCO}}_{3}{}^{-}[/latex], so H₂CO₃ is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and [latex]{\text{HCO}}_{3}{}^{-}[/latex] produced by ionization of H₂CO₃. (2) Then we determine the concentration of [latex]{\text{CO}}_{3}{}^{2-}[/latex] in a solution with the concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and [latex]{\text{HCO}}_{3}{}^{-}[/latex] determined in (1). To summarize:

Step 1: Determine the concentrations of [latex]\text{H}_{3}\text{O}^{+}[/latex] and [latex]\text{HCO}_{3}^{-}[/latex]

As for the ionization of any other weak acid:

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

An abbreviated table of changes and concentrations shows:

Substituting the equilibrium concentrations into the equilibrium gives us:

[latex]{K}_{{\text{H}}_{2}{\text{CO}}_{3}}=\dfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{HCO}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\dfrac{\left(x\right)\left(x\right)}{0.033-x}=4.3\times {10}^{-7}[/latex]

Assuming x << 0.033 and solving the simplified equation yields

[latex]x=1.2\times {10}^{-4}[/latex]

The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:

[latex]\left[{\text{H}}_{2}{\text{CO}}_{3}\right]=0.033M[/latex]

[latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\left[{\text{HCO}}_{3}{}^{-}\right]=1.2\times {10}^{-4}M[/latex]

Step 2: Determine the concentration of [latex]\text{CO}_{3}^{2-}[/latex] in a solution at equilibrium with [latex]\left[\text{H}_{3}\text{O}^{+}\right][/latex] and [latex]\left[\text{HCO}_{3}^{-}\right][/latex] both equal to 1.2 × 10⁻⁴ M.

[latex]{\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)[/latex]

[latex]{K}_{{\text{HCO}}_{3}{}^{-}}=\dfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}=\dfrac{\left(1.2\times {10}^{-4}\right)\left[{\text{CO}}_{3}{}^{2-}\right]}{1.2\times {10}^{-4}}[/latex]

[latex]\left[{\text{CO}}_{3}{}^{2-}\right]=\dfrac{\left(4.7\times {10}^{-11}\right)\left(1.2\times {10}^{-4}\right)}{1.2\times {10}^{-4}}=4.7\times {10}^{-11}M[/latex]

To summarize: In part 1 of this example, we found that the H₂CO₃ in a 0.033-M solution ionizes slightly and at equilibrium [H₂CO₃] = 0.033 M; [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 1.2 [latex]\times[/latex] 10⁻⁴; and [latex]\left[{\text{HCO}}_{3}{}^{-}\right]=1.2\times {10}^{-4}M[/latex]. In part 2, we determined that [latex]\left[{\text{CO}}_{3}{}^{2-}\right]=4.7\times {10}^{-11}M[/latex].

Check Your Learning

The concentration of H₂S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex], [HS⁻], and [S²⁻] in the solution:

[latex]\begin{array}{lll} {}{\text{H}}_{2}\text{S}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HS}}^{-}\left(aq\right)&{}&{K}_{\text{a}1}=1.0\times {10}^{-7}\\{\text{HS}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)&{}&{K}_{\text{a}2}=1.0\times {10}^{-19}\end{array}[/latex]

Show Solution

A triprotic acid is an acid that has three dissociable H atoms. Phosphoric acid is a typical example:

[latex]\begin{array}{l}\text{First ionization: }{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{H}}_{2}{\text{PO}}_{4}{}^{-}\left(aq\right)\qquad{K}_{\text{a}1}=7.5\times {10}^{-3}\\ \text{Second ionization: }{\text{H}}_{2}{\text{PO}}_{4}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HPO}}_{4}{}^{2-}\left(aq\right)\qquad{K}_{\text{a}2}=6.3\times {10}^{-8}\\ \text{Third ionization: }{\text{HPO}}_{4}{}^{2-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{PO}}_{4}{}^{3-}\left(aq\right)\qquad{K}_{\text{a}3}=3.6\times {10}^{-13}\end{array}[/latex]

As with the diprotic acids, examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10⁵ to 10⁶.

This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H₃PO₄ complicated. However, because the successive ionization constants differ by a factor of 10⁵ to 10⁶, the calculations can be broken down into a series of parts similar to those for diprotic acids.

Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:

[latex]{\text{H}}_{2}\text{O}\left(l\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\rightleftharpoons {\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\qquad{K}_{\text{b}1}=2.1\times{10}^{-4}[/latex]

[latex]{\text{H}}_{2}\text{O}\left(l\right)+{\text{HCO}}_{3}{}^{-}\left(aq\right)\rightleftharpoons {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\qquad{K}_{\text{b}2}=2.3\times{10}^{-8}[/latex]

You can view the transcript for “Acids and Bases, pH and pOH” here (opens in new window).

Key Concepts and Summary

An acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.

Try It

Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H₂CO₃, a diprotic acid: [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex], [OH⁻], [H₂CO₃], [latex]\left[{\text{HCO}}_{3}{}^{-}\right][/latex], [latex]\left[{\text{CO}}_{3}{}^{2-}\right]?[/latex] No calculations are needed to answer this question.
Calculate the concentration of each species present in a 0.050-M solution of H₂S.
Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C₆H₄(CO₂H)₂.

Show Selected Solutions

Try It

Salicylic acid, HOC₆H₄CO₂H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
1. Both functional groups of salicylic acid ionize in water, with K_a = 1.0 [latex]\times[/latex] 10⁻³ for the—CO₂H group and 4.2 [latex]\times[/latex] 10⁻¹³ for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).
2. Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH₃CO₂C₆H₄CO₂H. The −CO₂H functional group is still present, but its acidity is reduced, K_a = 3.0 [latex]\times[/latex] 10⁻⁴. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
3. Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:
  [latex]{\text{CH}}_{3}{\text{CO}}_{2}{\text{C}}_{6}{\text{H}}_{4}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{HOC}}_{6}{\text{H}}_{4}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)[/latex]
  1. Which of the acids salicylic acid or acetic acid produces more hydronium ions in solution such a solution?
  2. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?
The ion HTe⁻ is an amphiprotic species; it can act as either an acid or a base.
1. What is K_a for the acid reaction of HTe⁻ with H₂O?
2. What is K_b for the reaction in which HTe⁻ functions as a base in water?
3. Demonstrate whether or not the second ionization of H₂Te can be neglected in the calculation of [HTe⁻] in a 0.10 M solution of H₂Te.

Show Solution to Question 2

Part A

As an acid:

[latex]\begin{array}{l}\\ {\text{HTe}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{Te}}^{2-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)\\ {K}_{\text{a2}}=\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{-}\right]}=1\times {10}^{-5}\end{array}[/latex]

Part B

As a base:

[latex]\begin{array}{l}\\ {\text{HTe}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{Te}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ {K}_{\text{b}}=\frac{\left[{\text{H}}_{2}\text{Te}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HTe}}^{-}\right]}=\frac{{K}_{\text{w}}}{{K}_{\text{a1}}}=\frac{1.0\times {10}^{-14}}{2.3\times {10}^{-3}}=4.3\times {10}^{-12}\end{array}[/latex]

Part C

The reactions and ionization constants are:

[latex]\begin{array}{l}{ }{\text{H}}_{2}\text{Te}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HTe}}^{-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right){K}_{\text{a1}}=2.3\times {10}^{-3}\\ {\text{HTe}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{Te}}^{2-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right){K}_{\text{a2}}=1\times {10}^{-5}\end{array}[/latex]

As a general rule, if the first ionization constant is larger than the second by a factor of at least 20, then the second ionization can be neglected. Since [latex]{K}_{\text{a1}}[/latex] is 230-times larger than [latex]{K}_{\text{a2}}[/latex], the assumption should hold true for HTe⁻. To test the assumptions, find [HTe⁻] from the first reaction. The equilibrium expression for this reaction is [latex]{K}_{\text{a1}}=\frac{\left[{\text{HTe}}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{2}\text{Te}\right]}=2.3\times{10}^{-3}[/latex].

The initial and equilibrium concentrations for this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.10 − x) ≈ 0.10, gives [latex]\frac{\left[{\text{HTe}}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{2}\text{Te}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.10}=2.3\times {10}^{-3}[/latex].

Solving for x gives 0.0152 M. Because this value is 15% of 0.10 M, our assumption is incorrect. Therefore, use the quadratic formula. Using the data gives the quadratic equation:

x² + 2.3 [latex]\times[/latex] 10⁻³x − 2.3 [latex]\times[/latex] 10⁻⁴ = 0

Using the quadratic formula gives (a = 1, b = 2.3 [latex]\times[/latex] 10⁻³, and c = −2.3 [latex]\times[/latex] 10⁻⁴)

[latex]\begin{array}{ll}\hfill x& =\frac{-b\pm \sqrt{{b}^{\text{2+}}-4ac}}{2a}=\frac{-\left(2.3\times {10}^{-3}\right)\pm \sqrt{{\left(2.3\times {10}^{-3}\right)}^{\text{2+}}-4\left(1\right)\left(\text{-2.3}\times {10}^{-4}\right)}}{2\left(1\right)}\hfill \\ & =\frac{-\left(2.3\times {10}^{-3}\right)\pm \left(0.0304\right)}{2}=0.0141M\text{(positive root)}\hfill \end{array}[/latex]

Thus [HTe⁻] = 0.014 M. For the second ionization, [latex]{K}_{\text{a2}}=\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{-}\right]}=1\times{10}^{-5}[/latex].

The initial and equilibrium concentrations for this system can be written as follows:

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.0140 − x) ≈ and (0.0141 + x) ≈ 0.0141, gives:

[latex]\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{-}\right]}=\frac{\left(x\right)\left(0.0141+x\right)}{\left(0.0141-x\right)}\approx \frac{\left(x\right)\left(0.0141\right)}{0.0141}=1\times{10}^{-5}[/latex]

Solving for x gives 1 [latex]\times[/latex] 10⁻⁵M. Therefore, compared with 0.014 M, this value is negligible (0.071%).

Glossary

diprotic acid: acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps

diprotic base: base capable of accepting two protons. The protons are accepted in two steps

monoprotic acid: acid containing one ionizable hydrogen atom per molecule

stepwise ionization: process in which an acid is ionized by losing protons sequentially

triprotic acid: acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps

Module 14: Acid-Based Equilibria

Learning Outcomes

Example 1: Ionization of a Diprotic Acid

Step 1: Determine the concentrations of [latex]\text{H}_{3}\text{O}^{+}[/latex] and [latex]\text{HCO}_{3}^{-}[/latex]

Step 2: Determine the concentration of [latex]\text{CO}_{3}^{2-}[/latex] in a solution at equilibrium with [latex]\left[\text{H}_{3}\text{O}^{+}\right][/latex] and [latex]\left[\text{HCO}_{3}^{-}\right][/latex] both equal to 1.2 × 10⁻⁴ M.

Check Your Learning

Key Concepts and Summary

Try It

Try It

Part A

Part B

Part C

Glossary

Candela Citations

Module 14: Acid-Based Equilibria