\[\frac{\left ( 1.2gNaH \right )}{2.0L}\left ( \frac{molNaH}{24.0g} \right )\left ( \frac{molOH^-}{molNaH} \right )=0.025M OH^- \\ For example, if the answer is 1 x 10 -5, type "1e-5". You will learn how to calculate the isoelectric point, and the effects of pH on the amino acid's overall charge. \(x\) is given by the quadratic equation: \[x=\dfrac{b\sqrt{b^{2+}4ac}}{2a} \nonumber \]. Here are the steps to calculate the pH of a solution: Let's assume that the concentration of hydrogen ions is equal to 0.0001 mol/L. In strong bases, the relatively insoluble hydrated aluminum hydroxide, \(\ce{Al(H2O)3(OH)3}\), is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion: \[[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber \]. K a values can be easily looked up online, and you can find the pKa using the same operation as for pH if it is not listed as well. Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. So there is a second step to these problems, in that you need to determine the ionization constant for the basic anion of the salt. So for this problem, we It's easy to do this calculation on any scientific . At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). number compared to 0.20, 0.20 minus x is approximately The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100: \[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}100\% \label{PercentIon} \]. For example CaO reacts with water to produce aqueous calcium hydroxide. In this case, protons are transferred from hydronium ions in solution to \(\ce{Al(H2O)3(OH)3}\), and the compound functions as a base. Use this equation to calculate the percent ionization for a 1x10-6M solution of an acid with a Ka = 1x10-4M, and discuss (explain) the answer. High electronegativities are characteristic of the more nonmetallic elements. Determine x and equilibrium concentrations. Because water is the solvent, it has a fixed activity equal to 1. Posted 2 months ago. have from our ICE table. 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Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures. The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid. also be zero plus x, so we can just write x here. The product of these two constants is indeed equal to \(K_w\): \[K_\ce{a}K_\ce{b}=(1.810^{5})(5.610^{10})=1.010^{14}=K_\ce{w} \nonumber \]. The extent to which an acid, \(\ce{HA}\), donates protons to water molecules depends on the strength of the conjugate base, \(\ce{A^{}}\), of the acid. So we plug that in. We said this is acceptable if 100Ka <[HA]i. ICE table under acidic acid. 1.2 g sodium hydride in two liters results in a 0.025M NaOH that would have a pOH of 1.6. Step 1: Determine what is present in the solution initially (before any ionization occurs). The following data on acid-ionization constants indicate the order of acid strength: \(\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}\), \[ \begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.810^{5} \\[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.610^{-4} \\[4pt] \ce{HSO4-}(aq)+\ce{H2O}(l) &\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.210^{2} \end{aligned} \nonumber \]. Some anions interact with more than one water molecule and so there are some polyprotic strong bases. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. pH = pKa + log_ {10}\dfrac { [A^ {-}]} { [HA]} pH =pK a+log10[H A][A] This means that given an acid's pK a and the relative concentration of anion and "intact" acid, you can determine the pH. )%2F16%253A_AcidBase_Equilibria%2F16.06%253A_Weak_Acids, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Calculation of Percent Ionization from pH, Equilibrium Concentrations in a Solution of a Weak Acid, Equilibrium Concentrations in a Solution of a Weak Base. so \[\large{K'_{b}=\frac{10^{-14}}{K_{a}}}\], \[[OH^-]=\sqrt{K'_b[A^-]_i}=\sqrt{\frac{K_w}{K_a}[A^-]_i} \\ pH is a standard used to measure the hydrogen ion concentration. We write an X right here. Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\PageIndex{7}\)). What is the pH if 10.0 g Acetic Acid is diluted to 1.00 L? Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. was less than 1% actually, then the approximation is valid. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate \(\ce{Al(H2O)3(OH)3}\), is reflected in its solubility in both strong acids and strong bases. We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.810^{3}\:M. \\[4pt] &=9.810^{3}\:M \end{align*} \nonumber \]. More about Kevin and links to his professional work can be found at www.kemibe.com. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reactionso we do not need to consider itwhen setting up the ICE table. \[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}100 \nonumber \]. find that x is equal to 1.9, times 10 to the negative third. \[\begin{align} x^2 & =K_a[HA]_i \nonumber \\ x & =\sqrt{K_a[HA]_i} \nonumber \\ [H^+] & =\sqrt{K_a[HA]_i}\end{align}\]. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation. Given: pKa and Kb Asked for: corresponding Kb and pKb, Ka and pKa Strategy: The constants Ka and Kb are related as shown in Equation 16.6.10. It's going to ionize Note complete the square gave a nonsense answer for row three, as the criteria that [HA]i >100Ka was not valid. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Figure \(\PageIndex{1}\) lists several strong bases. And for the acetate The base ionization constant Kb of dimethylamine ( (CH3)2NH) is 5.4 10 4 at 25C. Calculate the concentration of all species in 0.50 M carbonic acid. there's some contribution of hydronium ion from the Generically, this can be described by the following reaction equations: \[H_2A(aq) + H_2O)l) \rightleftharpoons HA^- (aq) + H_3O^+(aq) \text{, where } K_{a1}=\frac{[HA^-][H_3O^+]}{H_2A} \], \[ HA^-(aq) +H_2O(l) \rightleftharpoons A^{-2}(aq) + H_3O^+(aq) \text{, where } K_{a2}=\frac{[A^-2][H_3O^+]}{HA^-}\]. What is the pH of a solution made by dissolving 1.2g lithium nitride to a total volume of 2.0 L? For hydroxide, the concentration at equlibrium is also X. From that the final pH is calculated using pH + pOH = 14. The equilibrium concentration of hydronium ions is equal to 1.9 times 10 to negative third Molar. \[\frac{\left ( 1.2gLi_3N\right )}{2.0L}\left ( \frac{molLi_3N}{34.83g} \right )\left ( \frac{3molOH^-}{molLi_3N} \right )=0.0517M OH^- \\ pOH=-log0.0517=1.29 \\ pH = 14-1.29 = 12.71 \nonumber \], \[pH=14+log(\frac{\left ( 1.2gLi_3N\right )}{2.0L}\left ( \frac{molLi_3N}{34.83g} \right )\left ( \frac{3molOH^-}{molLi_3N} \right )) = 12.71 \nonumber\]. A list of weak acids will be given as well as a particulate or molecular view of weak acids. Ninja Nerds,Join us during this lecture where we have a discussion on calculating percent ionization with practice problems! Ka values for many weak acids can be obtained from table 16.3.1 There are two cases. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, \(\ce{O2S(OH)2}\), sulfurous acid, \(\ce{OS(OH)2}\), nitric acid, \(\ce{O2NOH}\), perchloric acid, \(\ce{O3ClOH}\), aluminum hydroxide, \(\ce{Al(OH)3}\), calcium hydroxide, \(\ce{Ca(OH)2}\), and potassium hydroxide, \(\ce{KOH}\): If the central atom, E, has a low electronegativity, its attraction for electrons is low. of hydronium ion and acetate anion would both be zero. Our goal is to solve for x, which would give us the The equilibrium concentration of hydronium would be zero plus x, which is just x. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Calculate the pH of a solution prepared by adding 40.00mL of 0.237M HCl to 75.00 mL of a 0.133M solution of NaOH. Therefore, using the approximation This is the percentage of the compound that has ionized (dissociated). It you know the molar concentration of an acid solution and can measure its pH, the above equivalence allows . pH depends on the concentration of the solution. The equilibrium concentration Next, we can find the pH of our solution at 25 degrees Celsius. The second type of problem is to predict the pH or pOH for a weak base solution if you know Kb and the initial base concentration. Weak bases give only small amounts of hydroxide ion. the negative third Molar. Rule of Thumb: If \(\large{K_{a1}>1000K_{a2}}\) you can ignore the second ionization's contribution to the hydronium ion concentration, and if \([HA]_i>100K_{a1}\) the problem becomes fairly simple. If, for example, you have a 0.1 M solution of formic acid with a pH of 2.5, you can substitute this value into the pH equation: [H+] = 1 102.5 = 0.00316 M = 3.16 10-3 mol/L = 3.16 mmol/L. To solve, first determine pKa, which is simply log 10 (1.77 10 5) = 4.75. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H? You will want to be able to do this without a RICE diagram, but we will start with one for illustrative purpose. We will also discuss zwitterions, or the forms of amino acids that dominate at the isoelectric point. In one mixture of NaHSO4 and Na2SO4 at equilibrium, \(\ce{[H3O+]}\) = 0.027 M; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\). Then use the fact that the ratio of [A ] to [HA} = 1/10 = 0.1. pH = 4.75 + log 10 (0.1) = 4.75 + (1) = 3.75. First calculate the hydroxylammonium ionization constant, noting \(K'_aK_b=K_w\) and \(K_b = 8.7x10^{-9}\) for hydroxylamine. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\). Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. So we can go ahead and rewrite this. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber \], \[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.010^{3})(2.510^{3})}{0.050}=2.510^{4} \nonumber \]. In column 2 which was the limit, there was an error of .5% in percent ionization and the answer was valid to one sig. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A is its conjugate base, is as follows: HA ( aq) + H2O ( l) H3O + ( aq) + A ( aq) The equilibrium constant for this dissociation is as follows: K = [H3O +][A ] [H2O][HA] You can get Kb for hydroxylamine from Table 16.3.2 . \[[H^+]=\sqrt{K'_a[BH^+]_i}=\sqrt{\frac{K_w}{K_b}[BH^+]_i} \\ We can use pH to determine the Ka value. Check Your Learning Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89. got us the same answer and saved us some time. There are two basic types of strong bases, soluble hydroxides and anions that extract a proton from water. In section 15.1.2.2 we discussed polyprotic acids and bases, where there is an equilbiria existing between the acid, the acid salts and the salts. Consider the ionization reactions for a conjugate acid-base pair, \(\ce{HA A^{}}\): with \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\). This gives an equilibrium mixture with most of the base present as the nonionized amine. }{\le} 0.05 \nonumber \], \[\dfrac{x}{0.50}=\dfrac{7.710^{2}}{0.50}=0.15(15\%) \nonumber \]. However, that concentration The equilibrium constant for an acid is called the acid-ionization constant, Ka. And if x is a really small What is the pH of a solution made by dissolving 1.21g calcium oxide to a total volume of 2.00 L? If the percent ionization Formula to calculate percent ionization. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. So that's the negative log of 1.9 times 10 to the negative third, which is equal to 2.72. These acids are completely dissociated in aqueous solution. The lower the pH, the higher the concentration of hydrogen ions [H +]. There are two types of weak acid calculations, and these are analogous to the two type of equilibrium calculations we did in sections 15.3 and 15.4. The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. just equal to 0.20. down here, the 5% rule. \[\begin{align}NaH(aq) & \rightarrow Na^+(aq)+H^-(aq) \nonumber \\ H^-(aq)+H_2O(l) &\rightarrow H_2(g)+OH^-(aq) \nonumber \\ \nonumber \\ \text{Net} & \text{ Equation} \nonumber \\ \nonumber \\ NaH(aq)+H_2O(l) & \rightarrow Na^+(aq) + H_2(g)+OH^-(aq) \end{align}\]. If \(\ce{A^{}}\) is a strong base, any protons that are donated to water molecules are recaptured by \(\ce{A^{}}\). Am I getting the math wrong because, when I calculated the hydronium ion concentration (or X), I got 0.06x10^-3. Let's go ahead and write that in here, 0.20 minus x. The larger the \(K_a\) of an acid, the larger the concentration of \(\ce{H3O+}\) and \(\ce{A^{}}\) relative to the concentration of the nonionized acid, \(\ce{HA}\). However, if we solve for x here, we would need to use a quadratic equation. You can calculate the pH of a chemical solution, or how acidic or basic it is, using the pH formula: pH = -log 10 [H 3 O + ]. This is [H+]/[HA] 100, or for this formic acid solution. What is important to understand is that under the conditions for which an approximation is valid, and how that affects your results. the balanced equation. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution. You can check your work by adding the pH and pOH to ensure that the total equals 14.00. concentrations plugged in and also the Ka value. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, \(\ce{[C8H10N4O2H+]}\) = 5.0 103 M, and [OH] = 2.5 103 M? In this lesson, we will calculate the acid ionization constant, describe its use, and use it to understand how different acids have different strengths. ( K a = 1.8 1 0 5 ). A stronger base has a larger ionization constant than does a weaker base. Because water is the solvent, it has a fixed activity equal to 1. In condition 1, where the approximation is valid, the short cut came up with the same answer for percent ionization (to three significant digits). Muscles produce lactic acid, CH3CH (OH)COOH (aq) , during exercise. Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, \(\ce{HCN}\) or \(\ce{NH4+}\). For the generic reaction of a strong acid Ka is a large number meaning it is a product favored reaction that goes to completion and we use a one way arrow. The strengths of Brnsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. pH + pOH = 14.00 pH + pOH = 14.00. For example, the oxide ion, O2, and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water: \[\ce{O^2-}(aq)+\ce{H2O}(l)\ce{OH-}(aq)+\ce{OH-}(aq) \nonumber \], \[\ce{NH2-}(aq)+\ce{H2O}(l)\ce{NH3}(aq)+\ce{OH-}(aq) \nonumber \]. Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x << 0.534\), which permits us to simplify the denominator term as \((0.534 x) = 0.534\). { "16.01:_Br\u00f8nsted-Lowry_Concept_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.02:_Water_and_the_pH_Scale" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.03:_Equilibrium_Constants_for_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.04:_Acid-Base_Properties_of_Salts" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.05:_Acid-Base_Equilibrium_Calculations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.06:_Molecular_Structure,_Bonding,_and_Acid-Base_Behavior" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.07:_Lewis_Concept_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:General_Information" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Review" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Intermolecular_Forces_and_Liquids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Solids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13:_Solutions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "14:_Rates_of_Chemical_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15:_Equilibria" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16:_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "17:_Aqueous_Equilibria" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "18:_Entropy_and_Free_Energy" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "19:_Electron_Transfer_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "20:_Coordination_Chemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "21:_Nuclear_Chemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Appendix_1:_Google_Sheets" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "authorname:belfordr", "hypothesis:yes", "showtoc:yes", "license:ccbyncsa", "licenseversion:40" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FUniversity_of_Arkansas_Little_Rock%2FChem_1403%253A_General_Chemistry_2%2FText%2F16%253A_Acids_and_Bases%2F16.05%253A_Acid-Base_Equilibrium_Calculations, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), 16.6: Molecular Structure, Bonding, and Acid-Base Behavior, status page at https://status.libretexts.org, Type2: Calculate final pH from initial concentration and K. 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