Question: Is B2 2-a Paramagnetic or Diamagnetic ? Type Formula K sp; Bromides : PbBr 2: 6.3 x 10-6: AgBr: 3.3 x 10-13: Carbonates : BaCO 3: 8.1 x 10-9: CaCO 3: 3.8 x 10-9: CoCO 3: 8.0 x 10-13: CuCO 3: 2.5 x 10-10: FeCO 3: 3.5 x 10-11: PbCO 3: 1.5 x 10-13: MgCO 3: 4.0 x 10-5: MnCO 3: 1.8 x 10-11: NiCO 3: 6.6 x 10-9: Ag 2 CO 3: 8.1 x 10-12: ZnCO 3: 1.5 x 10-11: Chlorides When we talk about acid and base reactions, reactivity (and acidity and basicity) is all relative. Therefore, [OH-] = 0.05 M. Since the concentration of OH- is known, the pOH value is more useful. Direct link to Lloyd Succes's post Starting from 7:53, the p, Posted 8 years ago. Note that ammonia and most organic bases release OH- ions due to hydrolysis, not dissociation. Solve the equation for Kb by dividing the Kw by the Ka. Question : Is MgBr2 ( Magnesium Bromide ) an ionic or covalent bond ? In order to degrade it, supercritical water is used to convert it to the syngas containing carbon monoxide, carbon dioxide, hydrogen and methane. - [Voiceover] Let's look General Chemistry Articles, Study Guides, and Practice Problems. Its concentration doesn't Let's write our equilibrium expression. Reactions of Acids and Bases In Analytical Chemistry. How to write an equilibrium expression for an acid-base reaction and how to evaluate the strength of an acid using Ka. pOH is calculated by the formula, The value for pH is needed and the relationship between pH and pOH is given by. So, pKa = -logKa and Ka =10-pka Although the pH of KOH or potassium hydroxide is extremely high (usually ranging from 10 to 13 in typical solutions), the exact value depends on the concentration of this strong base in water. a loan pair of electrons in the auction taking our proton, leaving those electrons behind. All right, so KA is Bern, Switzerland, 6-9 November 2001. And these electrons in green Acids and Bases: Calculating pH of a Strong Acid, Henderson-Hasselbalch Equation and Example, Acids and Bases: Titration Example Problem, Calculating the Concentration of a Chemical Solution. There is virtually no undissociated NaOH left in the solution as it is almost entirely ionized to ions. Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. So this is just a faster way of doing it and HCL is a strong acid. CDC - NIOSH Pocket Guide to Chemical Hazards, https://en.wikipedia.org/w/index.php?title=Potassium_hydroxide&oldid=1152475114, Short description is different from Wikidata, Pages using collapsible list with both background and text-align in titlestyle, Articles containing unverified chemical infoboxes, Creative Commons Attribution-ShareAlike License 3.0, This page was last edited on 30 April 2023, at 13:17. Using the equation \(K_{a2} = \dfrac{[H_3O^+][SO_4^2-^-]}{[HSO_4^-]}\), \(K_{a2} = 1.1 * 10^-2\), and an ICE Table to get \(x^2 + .0.0205x - 0.0001045 = 0\). That is not happening since the electron Hydrogen originally had stays with the atom it was bonded with. We get approximately 100% ionization, so everything turns into our products here and let's go ahead and write In this process, it is used to improve the yield of gas and amount of hydrogen in process. ThoughtCo, Aug. 29, 2022, thoughtco.com/calculating-ph-of-a-strong-base-problem-609588. 0000012605 00000 n stay mostly protonated. All right, so let's use 2020 0 obj <> endobj %PDF-1.4 % . So concentration of our products times concentration of CL minus, all over, right, we have HCL and we leave out water. At first glance this gives an equilibrium constant of, \[K=\frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}\]. Relative Strength of Acids & Bases. approximately 100% ionization, we have all products here. Calculate [OH] in a solution obtained by adding 1.70 g solid KOH to 1.00 L of 10.0 M NH. So we can define the percent ionization of a weak acidas, Let's calculate the % Ionization of 1.0M and 0.01 M Acetic acid (Ka=1.8x10-5). of our reactant, so we have HA over here, so we have HA. Answer = SCl6 is Polar What is polarand non-polar? 0000001614 00000 n Aside from these, the carbonates (CO32-) and bicarbonates (HCO3) are also considered weak bases. As someone who has to write intricate Excel worksheets for preparing buffers at our company, this program [Buffer Maker] seems amazing. Separation of the anodic and cathodic spaces in the electrolysis cell is essential for this process.[15]. This electron pair picks up Therefore, a monoprotic acid is an acid that can donate only one proton, while polyprotic acid can donate more than one proton. To do that you use. The base dissociation constant, or Kb, of sodium hydroxide, or NaOH, is approximately 1020. reaction coming to an equilibrium, you're gonna have a If H2O is present in a given equation will it ALWAYS be the BLB? Direct link to Maria's post Ka =(A-)*(H3O+)/(HA) This equation goes to completion because H2SO4 is a strong acid and \(K_{a1}>>1\). 0000002830 00000 n Here you are going to find accommodation mostly in bigger resorts. Potassium hydroxide is also known as caustic potash, lye, and potash lye. Water can actually . Because of their relatively higher solubility, calculating the concentration of, and therefore, the pH of their solutions, Ca(OH)2, Ba(OH)2, and Sr(OH)2 follow the same principles as the hydroxides of alkali metals. Requested URL: byjus.com/chemistry/potassium-hydroxide/, User-Agent: Mozilla/5.0 (iPhone; CPU iPhone OS 14_8_1 like Mac OS X) AppleWebKit/605.1.15 (KHTML, like Gecko) Version/14.1.2 Mobile/15E148 Safari/604.1. Lower molecular-weight alcohols such as methanol, ethanol, and propanols are also excellent solvents. Direct link to Mr Spock's post If you were to do the rec, Posted 8 years ago. Consider the generic acid HA which has the reaction and equilibrium constant of, \[HA(aq)+H_2O(l)H_3O^+(aq)+A^-(aq), \; K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\]. So this is the acid ionization constant or you might hear acid what that does to the KA, all right, a very small number divided by a very large number, this be our Bronsted-Lowry acid and this is going to be the acidic proton. Using pressure swing adsorption, we could separate various gases and then use power-to-gas technology to convert them to fuel. did concentration of reactants over the concentration of products), would that be your kb? Direct link to Andrew El-Alam's post Are there other noteworth, Posted 8 years ago. This material has bothoriginal contributions, and contentbuilt upon prior contributions of the LibreTexts Community and other resources,including but not limited to: This page titled 16.3: Equilibrium Constants for Acids and Bases is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford. So H3O plus, the conjugate acid and then A minus would be a base. Include the problem's values in the . Direct link to hannah's post The oxygen will have a +1, Posted 8 years ago. gives you a KA value, an ionization constant much less than one. In this weakened state, the hair is more easily cut by a razor blade. Potassium Hydroxide or KOH, is a strong base and will dissociate completely in water to K+ and OH-. We could solve all these problems using the techniques from the last chapter on equilbria, but instead we are going to develop short cut techniques, and identify when they are valid. Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004). If we know K we can determine the pH or hydronium ion concentration using a rice diagram where we start with pure acid and measure determine how much dissociates. As for pKb values of strong bases - NaOH, KOH, LiOH, Ca(OH)2 - pleas read the explanation in our FAQ section. Question = Is IF4-polar or nonpolar ? Strong acids have a large Ka and completely dissociate and so you just state the reaction goes to completion. We will use K (a or b) to represent the acid or base equilibrium constant and K' (b or a) to represent the equilibrium constant of the conjugate pair. KOH is an example of a strong base, which means it dissociates into its ions in aqueous solution. * Compiled from Appendix 5 Chem 1A, B, C Lab Manual and Zumdahl 6th Ed. Ka is only used for weak acids. If you think about the Nope! White Sand beach has become the most popular on the island and so attracts the largest amount of tourists. Therule of thumb we will for this approximation isif [B]initial>100Kbwe willignore xin the denominator and simplify the math, \[If \; [B]_{i}>100K_b\\ \; \\then \\ \; \\ [B]_{i}-x \approxeq[B]_{i} \\ \; \\ and \\ \; \\ K_b=\frac{x^2}{[B]_{i}}\], This allows us to avoid the quadratic equation and quickly solve for the hydroxideion concentration, \[ pOH=-log[OH^-] = -log\sqrt{K_b[B]_i}\], \[pH=14-pOH \\ \; \\ or \\ \; \\ pH=14+log\sqrt{K_b[B]_i}\]. For the generic acid: \[HA \rightleftharpoons H^+ + A^- \\ \; \\ K_a=\frac{[H^{+}][A^{-}]}{[HA]} \]. Ka of HC2H3O2 (or CH3COOH) = 1.8 x 10^-5 Ka of HCHO2 = 1.8 x 10^-4 Ka of HOCl = 3.5 x 10^-8 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kb of NH3 = 1.8 x 10^-5 Kb of HC2H5O2 = 6.4 x 10^-4 Kb of CH3NH2 = 4.4 x 10-4 Kb of CH3CH2NH2 = 5.6 x 10-4 Examples of Spectator Ions Br, Cl, K, Na Buffers contain significant amounts of what? our equilibrium expression. The reaction is especially useful for aromatic reagents to give the corresponding phenols.[14]. I think the point is the molecule's ability to either donate OH- or accept H+ because either of these will increase the pH . Question = Is if4+polar or nonpolar ? Here is how to perform the pH calculation. [12], About 121 g of KOH dissolve in 100 mL water at room temperature, which contrasts with 100 g/100 mL for NaOH. The saponification of fats with KOH is used to prepare the corresponding "potassium soaps", which are softer than the more common sodium hydroxide-derived soaps. Unlike strong bases, weak bases do not contain a hydroxide ion. Now lets look at 0.0001M Acetic Acid. Kb of KOH is oo, Ka2 of H2SO4 is 0.010. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. When we write the equilibrium expression, write KA is equal to the Then you use the quadratic equation to solve for X, to get \(x\) = 0.004226. There is significantly less information on Kb values for common strong bases than there is for the Ka for common strong acids. Besides, difference between pKa=-1 and pKa=-10 starts to influence calculation results for the solutions with very high ionic strengths, such calculations are dubious in any case. Here are some of the values of weak and strong acids and bases dissociation constants used by BATE when calculating pH of the solution and concetrations of all ions present. Strong acids donate protons very easily and so we can say this the stuff on the left to be the reactants. as a Bronsted-Lowry base and a lone pair of Question = Is C2F2polar or nonpolar ? If you need more details on strong and weak bases in organic chemistry, particularly how amines are used there, check out this post. KOH is also used for semiconductor chip fabrication (for example anisotropic wet etching). Potassium hydroxide is preferred over sodium hydroxide because its solutions are more conductive. Marked out of 10.00 Answer: P Flag question Question 27 Not yet answered Calculate the solubility (in mol/L and g/L) of PbSO4(s) endstream endobj 2041 0 obj<>/W[1 1 1]/Type/XRef/Index[28 1992]>>stream The potassium salts of carbonate, cyanide, permanganate, phosphate, and various silicates are prepared by treating either the oxides or the acids with KOH. Potassium hydroxide is an inorganic compound with the formula K OH, and is commonly called caustic potash.. What is the pH of a 0.05 M solution of Potassium Hydroxide? \[H_2A^- + H_2O HA^{-2} +H_3O^+ \; \; K_{a2}\] Like any equilibrium reaction, the larger the equilibrium constant, the more the reaction is shifted to the right. In many textbooks, the above values are never discussed and the author will often write this about the Ka of a strong acid: And the exact values are never discussed. lies to the left because acetic acid is not They can be further categorized into diprotic acids and triprotic acids, those which can donate two and three protons, respectively. How do you convert KA to KB? Answer: B2 2-is a Diamagnetic What is Paramagnetic and Diamagnetic ? General Kb expressions take the form Kb = [BH+][OH-] / [B]. I think that correlates to base strength Whats the relationship between Ka and pH? To log in and use all the features of Khan Academy, please enable JavaScript in your browser. \(H_2PO_4^- + H_2O \rightleftharpoons H_3O^+ + HPO_4^{2-}\), \(K_{a2} = [HPO_4^{2-}] = 6.3 \times 10^{-8}\). Once HA donates a proton, we're Helmenstine, Todd. Just like the strong acids, we recognize them by their ability to completely ionize in aqueous solutions. According to Brnsted and Lowry an acid is a proton donor and a base is a proton acceptor. 4H2O. We're gonna think about HCL is gonna function Dissociation can be also described by overall constants, as well as base dissociation constants or protonation constants. Because of its high affinity for water, KOH serves as a desiccant in the laboratory. Monoprotic acid/base corresponds to the donation/acceptance of, Polyprotic acid/base corresponds to the donation/acceptance of. extremely high value for your KA. Based on the Kb values, NH 3 is the strongest base, and it has a smaller p Kb value. The potassium ion is a spectator. The larger theKb, the stronger the base. Direct link to Hafsa Kaja Moinudeen's post In the acetic acid and wa, Posted 6 years ago. KOH, like NaOH, serves as a source of OH, a highly nucleophilic anion that attacks polar bonds in both inorganic and organic materials. 0000022537 00000 n change and so we leave, we leave H2O out of our [21] Entomologists wishing to study the fine structure of insect anatomy may use a 10% aqueous solution of KOH to apply this process.[22]. For an Acid Base Conjugate Pair. For example, in a process commonly referred to as "chemical cremation" or "resomation", potassium hydroxide hastens the decomposition of soft tissues, both animal and human, to leave behind only the bones and other hard tissues. process occurs 100%. These as well, are types of acid-base reactions where the base is the oxide ion (O2-) and water is the acid. You can find out more about our use, change your default settings, and withdraw your consent at any time with effect for the future by visiting Cookies Settings, which can also be found in the footer of the site. much, much, much greater than one here. This reaction is manifested by the "greasy" feel that KOH gives when touched; fats on the skin are rapidly converted to soap and glycerol. The procedure is very similar for weak bases. These electrons in green move off onto the oxygen right here, For example, ammonia is a weak base because it produces a hydroxide ion and its conjugate base ammonium ion: \[{K_{\rm{b}}}\;{\rm{ = }}\;\frac{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ }}}} \right]}}{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]}}\]. Because aggressive bases like KOH damage the cuticle of the hair shaft, potassium hydroxide is used to chemically assist the removal of hair from animal hides. Except where otherwise noted, data are given for materials in their, Catalyst for hydrothermal gasification process, Rmpp Chemie-Lexikon, 9th Ed. 0000014794 00000 n those electrons in red. Monoprotic acids are acids that can release only one proton per molecule and have one equivalence point. \[CH_3NH_2(aq) + H_2O(l) CH_3NH_3^+(aq)+OH^- (aq) \\ \\ K=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} = 5.0x10^{-4}\], \[A^-(aq) + H_2O(l) HA(aq) + OH^-(aq)\], \[K'_b=\frac{[HA][OH^-]}{[A^-]} \\ \text{ where} \; K_b \; \text{is the basic equilibrium constant of the conjugate base} \; A^- \; \text{of the weak acid HA}\]. We will now look at weak acids and bases, which do not completely dissociate, and use equilibrium constants to calculate equilibrium concentrations. Potassium hydroxide is an inorganic compound which is denoted by the chemical formula KOH. So lone pair of electrons on the oxygen pick up this proton leaving NaOH is a strong base that completely ionizes or dissociates into Na and OH-ions in a solution. Direct link to Yasmeen.Mufti's post Nope! All right, so this electron Answer : MgBr2 ( Magnesium Bromide ) is a Ionicbond What is che New Questions About Fantasy Football Symbols Answered and Why You Must Read Every Word of This Report. 0000001177 00000 n Similarly, a monoprotic base can only accept one proton, while a polyprotic base can accept more than one proton. A titration curve displays the multiple acid dissociation constants (\(K_a\)) as portrayed below. However, due to molecular forces, the value of the . The pH of a 0.05 M solution of Potassium Hydroxide is 12.7. Remember that diprotic acids donate protons stepwise and there is an amphoteric intermediate HA-, so in the reaction of a diprotic acid there are 5 chemical species, H2A, HA-, A-2, H+and OH-. We are not permitting internet traffic to Byjus website from countries within European Union at this time. What is the Kb of this base? Water can actually be a BLB or a BLA, it is "Amphoteric". "Acids and Bases - Calculating pH of a Strong Base." So it picked up a proton. So plus one formal charge on the oxygen and let's show those electrons in red. So let's go ahead and draw our products. this acid base reaction would just be to write out H2O plus HCL, gives us H3O plus, plus CL minus. Great question! Answer = if4+ isPolar What is polarand non-polar? For the definitions of Kan constants scroll down the page. \[HA^{2}- + H_2O A^{-3} +H_3O^+ \; \; K_{a3}\], Because pKa and pKb values are so small they are often recorded a pX values, where pX= -logX. pH calculator program - Base Acid Titration and Equilibria - dissociation constants pKa and pKb. To simplify the numbers, the negative logarithm ofKbis often used to get rid of the exponent. Generally speaking, these values are not used in calculations since, at common concentrations in chemistry, each substance is 100% dissociated. Direct link to Ayan Gangopadhyay's post Cl- is a weaker base beca, Posted 8 years ago. left with the conjugate base which is A minus. If you were to do the recipricol of the ka (i.e. Answer = C2Cl2 is Polar What is polarand non-polar? We would form the acetate anions. Molten KOH is used to displace halides and other leaving groups. And , Posted 8 years ago. startxref Figure\(\PageIndex{1}\): Relationship between acid or base strength and that of their conjugate base or acid. All right, so H3O plus, so let me go ahead and draw in hydronium. 0000006099 00000 n Once again let's follow For example, if a bottle reads 2.0MNaOH, it actually indicates that the concentration of hydroxide and sodium ions is 2.0Meach. Over here for our Certain species of gilled mushrooms, boletes, polypores, and lichens[23] are identifiable based on this color-change reaction. For each compound enter compound name (optional), concentration, volume and Ka/Kb or pKa/pKb values. \[H_3PO_4 + H_2O \rightleftharpoons H_3O^+ + H_2PO_4^- \nonumber \], \[K_{a1} = \dfrac{[H_3O^+][H_2PO_4^-]}{[H_3PO_4]} \nonumber \], (b) From part (a), \(x\) = [H2PO4-] = [H3O+] = 0.17 M. (c) To determine [H3O+] and [H2PO4-], it was assumed that the second ionization constant was insignificant. Calculate [OH] in a solution obtained by adding 1.50 g solid KOH to 1.00 L of 10.0 M NH. basic A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. Thewater is omittedfrom the equilibrium constant expression giving. So water is gonna function as a base that's gonna take a proton this acid base reaction would be just to write What is the pH after 0 mL of NaOH has been added? For every mole of KOH, there will be 1 mole of OH-, so the concentration of OH- will be the same as the concentration of KOH. If we know K we can determine the pOHfrom the rice diagram, and once we know pOH, we can determine pH (because pH + pOH =14). 0000002363 00000 n products we have H3O plus, so let's write the We will use K(a or b) to represent the acid or base equilibrium constant and K'(b or a) to represent the equilibrium constant of the conjugate pair. KOH reacts with carbon dioxide to give potassium bicarbonate: Historically, KOH was made by adding potassium carbonate to a strong solution of calcium hydroxide (slaked lime). equilibrium expression. reverse reaction, H3O plus donating a proton to A minus anion is not very good at accepting them. 0000003077 00000 n NaOH is classified as a strong base, which completely ionizes or dissociates in a solution into Na + and OH - ions. 2. When you visit the site, Dotdash Meredith and its partners may store or retrieve information on your browser, mostly in the form of cookies. The site owner may have set restrictions that prevent you from accessing the site. This idea of proton donor and proton acceptor is important in understanding monoprotic and polyprotic acids and bases because monoprotic corresponds to the transfer of one proton and polyprotic refers to the transfer of more than one proton. To do that you use, \[K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{2} \], Another necessary value is the \(pK_a\) value, and that is obtained through \(pK_a = {-logK_a}\), The procedure is very similar for weak bases. If you're seeing this message, it means we're having trouble loading external resources on our website. Note: If using scientific notation, use e for the scientific notation formatting (i.e. (Kb of NH is 1.80 10). Only the [OH] from the excess KOH is to be counted. As a result of the EUs General Data Protection Regulation (GDPR). Polyprotic Acids & Bases is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Christopher Spohrer & Zach Wyatt. If we think about Conjugate acids (cations) of strong bases are ineffective bases. going to be much less than one and that's how we recognize, that's one way to recognize a weak acid. Consider the generic acid HA which has the reaction and equilibrium constant of. behind on the oxygen. This is what we also saw when introducing thepHto quantify the acidity of the solution. And the exact values are never discussed. Let me show those electrons. This results in Acid Dissociation Constant (Ka) for aqueous systems: \[K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\]. then you would get back H2O and HA. In the last 2 videos, the arrow has gone from the water to the hydrogen but is it incorrect to have the arrow going in the opposite direction? In the acetic acid and water reaction, can the acetic acid grab a proton from water instead of donating it? Accessibility StatementFor more information contact us atinfo@libretexts.org. %%EOF This method of producing potassium hydroxide remained dominant until the late 19th century, when it was largely replaced by the current method of electrolysis of potassium chloride solutions. HSO (aq) + HCN (aq) HSO (aq) + CN (aq) A) HSO, CN B) HSO, HSO C) HSO, CN D) HCN, HSO B) HSO, HSO Consider the reaction below. So KA is equal to a concentration of H3O plus. So we're gonna make A minus. So either one is fine. And over here if you think Thus, the solution of 0.25 M Ca(OH)2 will contain 0.25 M Ca2+, and 0.50 M OH ions because each mole of Ca(OH)2 ionizes to one mole of Ca2+ and 2 moles of OH ions: All alkali metal and alkaline earth metal oxides, except BeO which is amphoteric, are basic as well because their reaction with water produces the corresponding hydroxide. Look at the KA value. Direct link to srhee98's post Around 5:30, it was expla, Posted 7 years ago. the A to make A minus. Water, H2O accepted a proton, so this is our Bronsted-Lowry base and then once H2O accepts a proton, we turn into hydronium H3O plus. " The following bases are listed as strong: In textbooks where this idea is discussed, one often sees this statement about the Kb of a strong base. The general equation of a weak base is. In the case of methanol the potassium methoxide (methylate) forms: That's gonna give this oxygen So these two electrons in red here are gonna pick up this We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. in and then for water, we leave water out of our "Acids and Bases - Calculating pH of a Strong Base." (Kb of NH is 1.80 10). KA which we call the acid, the acid ionization constant. Note, in this reaction the base removes a proton from the water and following the same logic for weak acids, we consider the water concentration to stay constant because only a small fraction of it reacts with the weak base, so: An example of the first type would be that of methyl amine, CH3NH2. { "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]()" }, 16.3: Equilibrium Constants for Acids and Bases, [ "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%2F16.03%253A_Equilibrium_Constants_for_Acids_and_Bases, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\).
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