Now solve for \(x\). 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. The pH of a solution is a measure of the hydrogen ions, or protons, present in that solution. You can get Kb for hydroxylamine from Table 16.3.2 . To check the assumption that \(x\) is small compared to 0.534, we calculate: \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.810^{3}}{0.534} \\[4pt] &=1.810^{2} \, \textrm{(1.8% of 0.534)} \end{align*} \nonumber \]. of hydronium ions, divided by the initial The first six acids in Figure \(\PageIndex{3}\) are the most common strong acids. 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. \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\), \(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\), \(K_a \times K_b = 1.0 \times 10^{14} = K_w \,(\text{at room temperature})\), \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}100\). This table shows the changes and concentrations: 2. The lower the pH, the higher the concentration of hydrogen ions [H +]. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. For an equation of the form. The equilibrium expression is: \[\ce{HCO2H}(aq)+\ce{H2O}(l)\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \]. - [Instructor] Let's say we have a 0.20 Molar aqueous There are two basic types of strong bases, soluble hydroxides and anions that extract a proton from water. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure \(\PageIndex{3}\) exhibit no observable acidic behavior when dissolved in water. arrow_forward Calculate [OH-] and pH in a solution in which the hydrogen sulfite ion, HSO3-, is 0.429 M and the sulfite ion is (a) 0.0249 M (b) 0.247 M (c) 0.504 M (d) 0.811 M (e) 1.223 M The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We also need to plug in the 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. Therefore, if we write -x for acidic acid, we're gonna write +x under hydronium. Whether you need help solving quadratic equations, inspiration for the upcoming science fair or the latest update on a major storm, Sciencing is here to help. This reaction has been used in chemical heaters and can release enough heat to cause water to boil. To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH: \[\ce{[H3O+]}=10^{2.34}=0.0046\:M \nonumber \]. 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 16.5.2 tabulates hydronium concentration for an acid with Ka=10-4 at three different concentrations, where [HA]i is greater than, less than or equal to 100 Ka. This gives: \[K_\ce{a}=1.810^{4}=\dfrac{x^{2}}{0.534} \nonumber \], \[\begin{align*} x^2 &=0.534(1.810^{4}) \\[4pt] &=9.610^{5} \\[4pt] x &=\sqrt{9.610^{5}} \\[4pt] &=9.810^{3} \end{align*} \nonumber \]. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. And when acidic acid reacts with water, we form hydronium and acetate. 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. Example \(\PageIndex{1}\): Calculation of Percent Ionization from pH, Example \(\PageIndex{2}\): The Product Ka Kb = Kw, The Ionization of Weak Acids and Weak Bases, Example \(\PageIndex{3}\): Determination of Ka from Equilibrium Concentrations, Example \(\PageIndex{4}\): Determination of Kb from Equilibrium Concentrations, Example \(\PageIndex{5}\): Determination of Ka or Kb from pH, Example \(\PageIndex{6}\): Equilibrium Concentrations in a Solution of a Weak Acid, Example \(\PageIndex{7}\): Equilibrium Concentrations in a Solution of a Weak Base, Example \(\PageIndex{8}\): Equilibrium Concentrations in a Solution of a Weak Acid, The Relative Strengths of Strong Acids and Bases, status page at https://status.libretexts.org, \(\ce{(CH3)2NH + H2O (CH3)2NH2+ + OH-}\), Assess the relative strengths of acids and bases according to their ionization constants, Rationalize trends in acidbase strength in relation to molecular structure, Carry out equilibrium calculations for weak acidbase systems, Show that the calculation in Step 2 of this example gives an, Find the concentration of hydroxide ion in a 0.0325-. equilibrium concentration of hydronium ions. so \[\large{K'_{b}=\frac{10^{-14}}{K_{a}}}\], \[[OH^-]=\sqrt{K'_b[A^-]_i}=\sqrt{\frac{K_w}{K_a}[A^-]_i} \\ Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. ***PLEASE SUPPORT US***PATREON | . 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. 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. Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid? On the other hand, when dissolved in strong acids, it is converted to the soluble ion \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion: \[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber \]. Legal. Creative Commons Attribution/Non-Commercial/Share-Alike. \(\ce{NH4+}\) is the slightly stronger acid (Ka for \(\ce{NH4+}\) = 5.6 1010). And for the acetate In a diprotic acid there are two species that can protonate water, the acid itself, and the ion formed when it loses one of the two acidic protons (the acid salt anion). The equilibrium concentration of hydronium ions is equal to 1.9 times 10 to negative third Molar. Show that the quadratic formula gives \(x = 7.2 10^{2}\). The last equation can be rewritten: [ H 3 0 +] = 10 -pH. down here, the 5% rule. Thus, the order of increasing acidity (for removal of one proton) across the second row is \(\ce{CH4 < NH3 < H2O < HF}\); across the third row, it is \(\ce{SiH4 < PH3 < H2S < HCl}\) (see Figure \(\PageIndex{6}\)). Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. Note, in the first equation we are removing a proton from a neutral molecule while in the second we are removing it from a negative anion. In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\PageIndex{7}\)). A $0.185 \mathrm{M}$ solution of a weak acid (HA) has a pH of $2.95 .$ Calculate the acid ionization constant $\left(K_{\mathrm{a}}\right)$ for th Transcript Hi in this question, we have to find out the percentage ionization of acid that is weak acid here now he is a weak acid, so it will dissociate into irons in the solution as this. What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base: \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber \]. Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base . The equilibrium concentration of hydronium would be zero plus x, which is just x. A check of our arithmetic shows that \(K_b = 6.3 \times 10^{5}\). Use the \(K_b\) for the nitrite ion, \(\ce{NO2-}\), to calculate the \(K_a\) for its conjugate acid. To understand when the above shortcut is valid one needs to relate the percent ionization to the [HA]i >100Ka rule of thumb. anion, there's also a one as a coefficient in the balanced equation. for initial concentration, C is for change in concentration, and E is equilibrium concentration. log of the concentration of hydronium ions. To figure out how much So we're going to gain in Only a small fraction of a weak acid ionizes in aqueous solution. Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1. Thus there is relatively little \(\ce{A^{}}\) and \(\ce{H3O+}\) in solution, and the acid, \(\ce{HA}\), is weak. 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. The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). \(x\) is given by the quadratic equation: \[x=\dfrac{b\sqrt{b^{2+}4ac}}{2a} \nonumber \]. going to partially ionize. So the equation 4% ionization is equal to the equilibrium concentration of hydronium ions, divided by the initial concentration of the acid, times 100%. the balanced equation showing the ionization of acidic acid. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a basethis is the case with Ca(OH)2 and KOH. So we can plug in x for the In an ICE table, the I stands \[\dfrac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right )=0.0216M OH^- \\[5pt] pOH=-\log0.0216=1.666 \\[5pt] pH = 14-1.666 = 12.334 \nonumber \], Note this could have been done in one step, \[pH=14+log(\frac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right)) = 12.334 \nonumber\]. If the percent ionization Steps for How to Calculate Percent Ionization of a Weak Acid or Base Step 1: Read through the given information to find the initial concentration and the equilibrium constant for the weak. We can use pH to determine the Ka value. A weak acid gives small amounts of \(\ce{H3O+}\) and \(\ce{A^{}}\). So this is 1.9 times 10 to The Ka value for acidic acid is equal to 1.8 times What is the pH of a solution made by dissolving 1.2g lithium nitride to a total volume of 2.0 L? Salts of a weak acid and a strong base form basic solutions because the conjugate base of the weak acid removes a proton from water. Calculate the concentration of all species in 0.50 M carbonic acid. There's a one to one mole ratio of acidic acid to hydronium ion. Water also exerts a leveling effect on the strengths of strong bases. autoionization of water. Am I getting the math wrong because, when I calculated the hydronium ion concentration (or X), I got 0.06x10^-3. 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. 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 negative third Molar. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water. This gives an equilibrium mixture with most of the base present as the nonionized amine. How can we calculate the Ka value from pH? Figure \(\PageIndex{3}\) lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. These acids are completely dissociated in aqueous solution. ICE table under acidic acid. So we plug that in. So for this problem, we Ka values for many weak acids can be obtained from table 16.3.1 There are two cases. The aciddissociation (or ionization) constant, Ka , of this acid is 8.40104 . Just like strong acids, strong Bases 100% ionize (KB>>0) and you solve directly for pOH, and then calculate pH from pH + pOH =14. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. \(K_a\) for \(\ce{HSO_4^-}= 1.2 \times 10^{2}\). Just like strong acids, strong Bases 100% ionize (K B >>0) and you solve directly for pOH, and then calculate pH from pH + pOH =14. The ionization constant of \(\ce{HCN}\) is given in Table E1 as 4.9 1010. Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber \]. %ionization = [H 3O +]eq [HA] 0 100% Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. we made earlier using what's called the 5% rule. pH is a standard used to measure the hydrogen ion concentration. The equilibrium constant for the ionization of a weak base, \(K_b\), is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. 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. 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". The \(\ce{Al(H2O)3(OH)3}\) compound thus acts as an acid under these conditions. Achieve: Percent Ionization, pH, pOH. Muscles produce lactic acid, CH3CH (OH)COOH (aq) , during exercise. giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.310^{5} \nonumber \]. As we begin solving for \(x\), we will find this is more complicated than in previous examples. quadratic equation to solve for x, we would have also gotten 1.9 water to form the hydronium ion, H3O+, and acetate, which is the At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \\[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \\[4pt] &=1.7710^{5} \end{align*} \nonumber \]. More about Kevin and links to his professional work can be found at www.kemibe.com. 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 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. That is, the amount of the acid salt anion consumed or the hydronium ion produced in the second step (below) is negligible and so the first step determines these concentrations. \(x\) is less than 5% of the initial concentration; the assumption is valid. reaction hasn't happened yet, the initial concentrations This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. 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] Would the proton be more attracted to HA- or A-2? 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 \(K_a\). (Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.) As in the previous examples, we can approach the solution by the following steps: 1. where the concentrations are those at equilibrium. What is the pH of a 0.50-M solution of \(\ce{HSO4-}\)? and you should be able to derive this equation for a weak acid without having to draw the RICE diagram. the balanced equation. The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). Check out the steps below to learn how to find the pH of any chemical solution using the pH formula. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds. Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root: Now determine the hydronium ion concentration and the pH: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.210^{2}\:M \\[4pt] &=7.210^{2}\:M \end{align*} \nonumber \], \[\mathrm{pH=log[H_3O^+]=log7.210^{2}=1.14} \nonumber \], \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.510^{4} \nonumber \]. Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) We also need to calculate Some of the acidic acid will ionize, but since we don't know how much, we're gonna call that x. Determine x and equilibrium concentrations. pH = 14+log\left ( \sqrt{\frac{K_w}{K_a}[A^-]_i} \right )\]. 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. 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. 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 + ]. in section 16.4.2.3 we determined how to calculate the equilibrium constant for the conjugate base of a weak acid. Legal. We can rank the strengths of acids by the extent to which they ionize in aqueous solution. 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. 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. This is similar to what we did in heterogeneous equilibiria where we omitted pure solids and liquids from equilibrium constants, but the logic is different (this is a homogeneous equilibria and water is the solvent, it is not a separate phase). The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). Next, we can find the pH of our solution at 25 degrees Celsius. \[ [H^+] = [HA^-] = \sqrt {K_{a1}[H_2A]_i} \\ = \sqrt{(4.5x10^{-7})(0.50)} = 4.7x10^{-4}M \nonumber\], \[[OH^-]=\frac{10^{-14}}{4.74x10^{-4}}=2.1x10^{-11}M \nonumber\], \[[H_2A]_e= 0.5 - 0.00047 =0.50 \nonumber\], \[[A^{-2}]=K_{a2}=4.7x10^{-11}M \nonumber\]. there's some contribution of hydronium ion from the This is all over the concentration of ammonia and that would be the concentration of ammonia at equilibrium is 0.500 minus X. What is the value of \(K_a\) for acetic 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. Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. where the concentrations are those at equilibrium. The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. The percent ionization of a weak acid, HA, is defined as the ratio of the equilibrium HO concentration to the initial HA concentration, multiplied by 100%. Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. 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%. pH=-log\sqrt{\frac{K_w}{K_b}[BH^+]_i}\]. of the acetate anion also raised to the first power, divided by the concentration of acidic acid raised to the first power. the equilibrium concentration of hydronium ions. So the Molars cancel, and we get a percent ionization of 0.95%. Another way to look at that is through the back reaction. Since \(\large{K_{a1}>1000K_{a2}}\) the acid salt anion \(HA^-\) and \(H_3O^+\) concentrations come from the first ionization. The chemical equation for the dissociation of the nitrous acid is: \[\ce{HNO2}(aq)+\ce{H2O}(l)\ce{NO2-}(aq)+\ce{H3O+}(aq). So the equation 4% ionization is equal to the equilibrium concentration Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{}}\) is the conjugate base of the acid \(\ce{HA}\), and the hydronium ion is the conjugate acid of water. What is Kb for NH3. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. H+ is the molarity. Therefore, using the approximation conjugate base to acidic acid. In these problems you typically calculate the Ka of a solution of known molarity by measuring it's pH. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09. And our goal is to calculate the pH and the percent ionization. Note this could have been done in one step So the Ka is equal to the concentration of the hydronium ion. In this problem, \(a = 1\), \(b = 1.2 10^{3}\), and \(c = 6.0 10^{3}\). 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. Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). to the first power, times the concentration Consider the ionization reactions for a conjugate acid-base pair, \(\ce{HA A^{}}\): with \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\). 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. Some anions interact with more than one water molecule and so there are some polyprotic strong bases. Formic acid, HCO2H, is the irritant that causes the bodys reaction to ant stings. Posted 2 months ago. This error is a result of a misunderstanding of solution thermodynamics. The reaction of a Brnsted-Lowry base with water is given by: \[\ce{B}(aq)+\ce{H2O}(l)\ce{HB+}(aq)+\ce{OH-}(aq) \nonumber \]. In this lesson, we will calculate the acid ionization constant, describe its use, and use it to understand how different acids have different strengths. Step 1: Determine what is present in the solution initially (before any ionization occurs). Weak acids are acids that don't completely dissociate in solution. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color): 2. So to make the math a little bit easier, we're gonna use an approximation. ), { "16.01:_Acids_and_Bases_-_A_Brief_Review" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.
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