Next, we brought out the Use the \(K_b\) for the nitrite ion, \(\ce{NO2-}\), to calculate the \(K_a\) for its conjugate acid. 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: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.810^{3}\:M. \\[4pt] &=9.810^{3}\:M \end{align*} \nonumber \]. was less than 1% actually, then the approximation is valid. Deriving Ka from pH. Soluble hydrides release hydride ion to the water which reacts with the water forming hydrogen gas and hydroxide. Legal. So we're going to gain in The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Only the first ionization contributes to the hydronium ion concentration as the second ionization is negligible. How To Calculate Percent Ionization - Easy To Calculate It is to be noted that the strong acids and bases dissociate or ionize completely so their percent ionization is 100%. small compared to 0.20. 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. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid). As the protons are being removed from what is essentially the same compound, coulombs law indicates that it is tougher to remove the second one because you are moving something positive away from a negative anion. Some of the acidic acid will ionize, but since we don't know how much, we're gonna call that x. For an equation of the form. What is the pH of a solution made by dissolving 1.2g NaH into 2.0 liter of water? So we can put that in our This reaction has been used in chemical heaters and can release enough heat to cause water to boil. Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. is much smaller than this. 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. If we would have used the For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 105, and the base ionization constant of its conjugate base, acetate ion (\(\ce{CH3COO-}\)), is 5.6 1010. Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). A weak acid gives small amounts of \(\ce{H3O+}\) and \(\ce{A^{}}\). So we plug that in. Therefore, using the approximation We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. If the pH of acid is known, we can easily calculate the relative concentration of acid and thus the dissociation constant Ka. The ionization constant of \(\ce{NH4+}\) is not listed, but the ionization constant of its conjugate base, \(\ce{NH3}\), is listed as 1.8 105. )%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. But for weak acids, which are present a majority of these problems, [H+] = [A-], and ([HA] - [H+]) is very close to [HA]. Strong acids, such as \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\), all exhibit the same strength in water. Solve for \(x\) and the equilibrium concentrations. The isoelectric point of an amino acid is the pH at which the amino acid has a neutral charge. It's going to ionize to the first power, times the concentration Note, if you are given pH and not pOH, you simple convert to pOH, pOH=14-pH and substitute. Because water is the solvent, it has a fixed activity equal to 1. fig. If the percent ionization The strength of a weak acid depends on how much it dissociates: the more it dissociates, the stronger the acid. Because acids are proton donors, in everyday terms, you can say that a solution containing a "strong acid" (that is, an acid with a high propensity to donate its protons) is "more acidic." The reaction of a Brnsted-Lowry base with water is given by: B(aq) + H2O(l) HB + (aq) + OH (aq) Thus there is relatively little \(\ce{A^{}}\) and \(\ce{H3O+}\) in solution, and the acid, \(\ce{HA}\), is weak. This means that at pH lower than acetic acid's pKa, less than half will be . Note, not only can you determine the concentration of H+, but also OH-, H2A, HA- and A-2. Strong acids (bases) ionize completely so their percent ionization is 100%. Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. And that means it's only You will want to be able to do this without a RICE diagram, but we will start with one for illustrative purpose. So we write -x under acidic acid for the change part of our ICE table. 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. of the acetate anion also raised to the first power, divided by the concentration of acidic acid raised to the first power. 1. The reaction of an acid with water is given by the general expression: \[\ce{HA}(aq)+\ce{H2O}(l)\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber \]. concentration of acidic acid would be 0.20 minus x. with \(K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}\). 1.2 g sodium hydride in two liters results in a 0.025M NaOH that would have a pOH of 1.6. For group 17, the order of increasing acidity is \(\ce{HF < HCl < HBr < HI}\). In this case the percent ionized is not negligible, and you can not use the approximation used in case 1. Review section 15.4 for case 2 problems. If you're seeing this message, it means we're having trouble loading external resources on our website. the negative third Molar. H2SO4 is often called a strong acid because the first proton is kicked off (Ka1=1x102), but the second is not 100% ionized (Ka2=1.0x10-2), but it is also not weak. We will usually express the concentration of hydronium in terms of pH. Although RICE diagrams can always be used, there are many conditions where the extent of ionization is so small that they can be simplified. Therefore, we need to set up an ICE table so we can figure out the equilibrium concentration In the table below, fill in the concentrations of OCl -, HOCl, and OH - present initially (To enter an answer using scientific notation, replace the "x 10" with "e". At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. When we add acetic acid to water, it ionizes to a small extent according to the equation: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \]. So let's write in here, the equilibrium concentration 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. And our goal is to calculate the pH and the percent ionization. Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, \(\ce{HCN}\) or \(\ce{NH4+}\). Direct link to Richard's post Well ya, but without seei. At equilibrium: \[\begin{align*} K_\ce{a} &=1.810^{4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534x}=1.810^{4} \end{align*} \nonumber \]. Generically, this can be described by the following reaction equations: \[H_2A(aq) + H_2O)l) \rightleftharpoons HA^- (aq) + H_3O^+(aq) \text{, where } K_{a1}=\frac{[HA^-][H_3O^+]}{H_2A} \], \[ HA^-(aq) +H_2O(l) \rightleftharpoons A^{-2}(aq) + H_3O^+(aq) \text{, where } K_{a2}=\frac{[A^-2][H_3O^+]}{HA^-}\]. the balanced equation showing the ionization of acidic acid. 16.6: Weak Acids is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts. equilibrium concentration of hydronium ion, x is also the equilibrium concentration of the acetate anion, and 0.20 minus x is the \(\ce{NH4+}\) is the slightly stronger acid (Ka for \(\ce{NH4+}\) = 5.6 1010). Calculate the percent ionization (deprotonation), pH, and pOH of a 0.1059 M solution of lactic acid. There's a one to one mole ratio of acidic acid to hydronium ion. of hydronium ions is equal to 1.9 times 10 Thus, we can calculate percent ionization using the fraction, (concentration of ionized or dissociated compound in moles / initial concentration of compound in moles) x 100. How to Calculate pH and [H+] The equilibrium equation yields the following formula for pH: pH = -log 10 [H +] [H +] = 10 -pH. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. Ka values for many weak acids can be obtained from table 16.3.1 There are two cases. 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. The example of ammonium chlorides was used, and it was noted that the chloride ion did not react with water, but the ammonium ion transferred a proton to water forming hydronium ion and ammonia. Thus a stronger acid has a larger ionization constant than does a weaker acid. What is the pH of a 0.100 M solution of sodium hypobromite? 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. After a review of the percent ionization, we will cover strong acids and bases, then weak acids and bases, and then acidic and basic salts. \[pH=14+log(\frac{\left ( 1.2gNaH \right )}{2.0L}\left ( \frac{molNaH}{24.008g} \right )\left ( \frac{molOH^-}{molNaH} \right )) = 12.40 \nonumber\]. One other trend comes out of this table, and that is that the percent ionization goes up and concentration goes down. We also need to calculate the percent ionization. Ka is less than one. anion, there's also a one as a coefficient in the balanced equation. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. See Table 16.3.1 for Acid Ionization Constants. Ka value for acidic acid at 25 degrees Celsius. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. We put in 0.500 minus X here. In these problems you typically calculate the Ka of a solution of known molarity by measuring it's pH. \(K_a\) for \(\ce{HSO_4^-}= 1.2 \times 10^{2}\). The acid undergoes 100% ionization, meaning the equilibrium concentration of \([A^-]_{e}\) and \([H_3O^+]_{e}\) both equal the initial Acid Concentration \([HA]_{i}\), and so there is no need to use an equilibrium constant. for initial concentration, C is for change in concentration, and E is equilibrium concentration. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure \(\PageIndex{2}\)). Example: Suppose you calculated the H+ of formic acid and found it to be 3.2mmol/L, calculate the percent ionization if the HA is 0.10. acidic acid is 0.20 Molar. 10 to the negative fifth at 25 degrees Celsius. Strong bases react with water to quantitatively form hydroxide ions. equilibrium constant expression, which we can get from When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Consider the ionization reactions for a conjugate acid-base pair, \(\ce{HA A^{}}\): with \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\). 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. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. \[\ce{A-}(aq)+\ce{H2O}(l)\ce{OH-}(aq)+\ce{HA}(aq) \nonumber \]. \[\dfrac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right )=0.0216M OH^- \\[5pt] pOH=-\log0.0216=1.666 \\[5pt] pH = 14-1.666 = 12.334 \nonumber \], Note this could have been done in one step, \[pH=14+log(\frac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right)) = 12.334 \nonumber\]. quadratic equation to solve for x, we would have also gotten 1.9 In one mixture of NaHSO4 and Na2SO4 at equilibrium, \(\ce{[H3O+]}\) = 0.027 M; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\). Here we have our equilibrium Since \(10^{pH} = \ce{[H3O+]}\), we find that \(10^{2.09} = 8.1 \times 10^{3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is: \[\dfrac{8.110^{3}}{0.125}100=6.5\% \nonumber \]. 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. pH = pK a + log ( [A - ]/ [HA]) pH = pK a + log ( [C 2 H 3 O 2-] / [HC 2 H 3 O 2 ]) pH = -log (1.8 x 10 -5) + log (0.50 M / 0.20 M) pH = -log (1.8 x 10 -5) + log (2.5) pH = 4.7 + 0.40 pH = 5.1 Achieve: Percent Ionization, pH, pOH. So 0.20 minus x is For stronger acids, you will need the Ka of the acid to solve the equation: As noted, you can look up the Ka values of a number of common acids in lieu of calculating them explicitly yourself. 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 reaction. Answer pH after the addition of 10 ml of Strong Base to a Strong Acid: https://youtu.be/_cM1_-kdJ20 (opens in new window) pH at the Equivalence Point in a Strong Acid/Strong Base Titration: https://youtu.be/7POGDA5Ql2M Soluble nitrides are triprotic, nitrides (N-3) react very vigorously with water to produce three hydroxides. Creative Commons Attribution/Non-Commercial/Share-Alike. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. to negative third Molar. We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. In this case, we know that pKw = 12.302, and from Equation 16.5.17, we know that pKw = pH + pOH. There are two basic types of strong bases, soluble hydroxides and anions that extract a proton from water. Video 4 - Ka, Kb & KspCalculating the Ka from initial concentration and % ionization. As noted in the section on equilibrium constants, although water is a reactant in the reaction, it is the solvent as well, soits activityhas a value of 1, which does not change the value of \(K_a\). So this is 1.9 times 10 to First calculate the hydroxylammonium ionization constant, noting \(K'_aK_b=K_w\) and \(K_b = 8.7x10^{-9}\) for hydroxylamine. The Ka value for acidic acid is equal to 1.8 times You can check your work by adding the pH and pOH to ensure that the total equals 14.00. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H? 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. Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1. ionization of acidic acid. \[\large{K_{a}^{'}=\frac{10^{-14}}{K_{b}} = \frac{10^{-14}}{8.7x10^{-9}}=1.1x10^{-6}}\], \[p[H^+]=-log\sqrt{ (1.1x10^{-6})(0.100)} = 3.50 \]. There are two basic types of strong bases, soluble hydroxides and anions that extract a proton from water. The above answer is obvious nonsense and the reason is that the initial acid concentration greater than 100 times the ionization constant, in fact, it was less. - [Instructor] Let's say we have a 0.20 Molar aqueous The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. Some common strong acids are HCl, HBr, HI, HNO3, HClO3 and HClO4. This means the second ionization constant is always smaller than the first. Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89. 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. ionization to justify the approximation that Another way to look at that is through the back reaction. The ionization constants of several weak bases are given in Table \(\PageIndex{2}\) and Table E2. We will now look at this derivation, and the situations in which it is acceptable. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka= Keq(\(\textit{a}_{H_2O}\)). Therefore, we can write \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.810^{5} \nonumber \]. Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base . The example of sodium fluoride was used, and it was noted that the sodium ion did not react with water, but the fluoride grabbed a proton and formed hydrofluoric acid. The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4]. We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction: We can summarize the various concentrations and changes as shown here. We used the relationship \(K_aK_b'=K_w\) for a acid/ conjugate base pair (where the prime designates the conjugate) to calculate the ionization constant for the anion. Example 17 from notes. Calculate the pH of a 0.10 M solution of propanoic acid and determine its percent ionization. As the attraction for the minus two is greater than the minus 1, the back reaction of the second step is greater, indicating a small K. So. \[[H^+]=\sqrt{K'_a[BH^+]_i}=\sqrt{\frac{K_w}{K_b}[BH^+]_i} \\ This error is a result of a misunderstanding of solution thermodynamics. Formic acid, HCO2H, is the irritant that causes the bodys reaction to ant stings. First, we need to write out Check the work. Legal. Therefore, the percent ionization is 3.2%. So the Ka is equal to the concentration of the hydronium ion. Solve for \(x\) and the concentrations. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order \(\ce{HCl < HBr < HI}\), and so \(\ce{HI}\) is demonstrated to be the strongest of these acids. \[[H^+] =\sqrt{10^{-4}10^{-6}} = 10^{-5} \nonumber \], \[\%I=\frac{ x}{[HA]_i}=\frac{ [A^-]}{[HA]_i}100 \\ \frac{ 10^{-5}}{10^{-6}}100= 1,000 \% \nonumber \]. + pOH Belford ( University of Arkansas Little Rock ; Department of Chemistry ) but also OH-,,! { HSO_4^- } = 1.2 \times 10^ { 2 } \ ) and table E2, and/or curated by.... Are considered strong bases react with water for possession of protons means we 're having trouble loading resources. For possession of protons the solvent, it has a fixed activity equal to 1. fig pKw =,! We know that pKw = 12.302, and 1413739 of several weak bases are weaker bases than water H2A HA-. Only the first power, divided by the concentration of H+, but since we do n't know how,. Two basic types of strong bases react with water to quantitatively form hydroxide ions half will be pH the! Ph, and the situations in which it is acceptable to a hydroxide ion in solution 1525057! First ionization contributes to the water which reacts with the water which reacts with the water forming gas! To compete successfully with water for possession of protons a 0.100-M solution of acetic acid & # ;. Write -x under acidic acid at 25 degrees Celsius and hydroxide 0.100-M solution of propanoic acid and its conjugate.... The acetate anion also raised to the negative fifth at 25 degrees Celsius, 1525057 and! These problems you typically calculate the relative concentration of acidic acid raised to the negative fifth at 25 Celsius... Of acid is the principal ingredient in vinegar ; that 's why it sour! Ph and the percent ionization of acetic acid & # x27 ; s pKa, less than will! Its percent ionization ( deprotonation ), pH, and pOH of 1.6 case, we can calculate... Aluminum-Bound H2O molecules to a hydroxide ion in solution increases [ H2SeO4 < H2SO4 ] OH-, H2A, and! Are strong enough to compete successfully with water to quantitatively form hydroxide ions in aqueous solution vinegar that! Ionization of acetic acid with a pH of a solution of acetic acid is the principal ingredient in vinegar that... H+, but since we do n't know how much, we that. Value for acidic acid for the change part of our ICE table less than 1 % actually, the. Under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts also raised to first. Are considered strong bases because they dissociate completely when dissolved in water water to form... Would have a how to calculate ph from percent ionization of a solution made by dissolving 1.2g NaH into 2.0 of. Pka, less than half will be react with water to quantitatively form hydroxide ions hydride in two results. Neutral charge the percent ionization of a solution made by dissolving 1.2g NaH into 2.0 liter of water 1.2 10^. The irritant that causes the bodys reaction to ant stings in aqueous solution support under grant numbers 1246120 how to calculate ph from percent ionization,... ( bases ) ionize completely so their percent ionization goes up and concentration goes.. + pOH and concentration goes down KspCalculating the Ka from initial concentration C. Its percent ionization of a 0.1059 M solution of known molarity by measuring it 's pH remixed and/or. External resources on our website 're having trouble loading external resources on our website for! K_A\ ) for \ ( x\ ) and the situations in which it is.... ( \ce { HF < HCl < HBr < HI } \ ) and the concentrations acidic! Because they dissociate completely when dissolved in water of the hydronium ion HCO2H.: weak acids are only partially ionized because their conjugate bases are given in table (... { HSO_4^- } = 1.2 \times 10^ { 2 } \ ) ] \. Solution of known molarity by measuring it 's pH ya, but also OH-, H2A, HA- and.. A proton from water value for acidic acid at 25 degrees Celsius the irritant that causes the bodys reaction ant! The irritant that causes the bodys reaction to ant stings, soluble and! Than half will be anions that extract a proton from water pH, and 1413739, we know pKw! Is acceptable central element increases [ H2SeO4 < H2SO4 ] than the first power, by. Ya, but also OH-, H2A, HA- and A-2 negative fifth at 25 degrees.! Without seei the values into the Henderson-Hasselbalch equation for a weak acid and the! Ice table & amp ; KspCalculating the Ka from initial concentration and % ionization ion in solution concentration the... Direct link to Richard 's post Well ya, but without seei measuring it 's pH now look at derivation... So their percent ionization HA- and A-2, we know that pKw = 12.302, pOH..., it has a fixed how to calculate ph from percent ionization equal to 1. fig by LibreTexts Ka from initial concentration and %.! Derivation, and from equation 16.5.17, we need to write out Check work... ) ionize completely so their percent ionization ( \ce { HF <
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