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Already registered? You guys said the concentration I should have found is 0.0126M. The term "pH" defines the quality of the water we drink, but do you know what it means? If you can access a quad equation solver on your personal electronic device or through the Internet, this is quick and painless. What percentage of the acid is dissociated? lessons in math, English, science, history, and more. succeed. (See any textbook on numerical computing for more on this and other metnods.). The "degree of dissociation" (denoted by \(\alpha\) of a weak acid is just the fraction, \[\alpha = \dfrac{[\ce{A^{-}}]}{C_a} \label{1-13}\]. Hence, there is no need for ICE tables. Explanation: For a hypothetical weak acid H A H + +A Ka = ( [H +][A] H A) where [H +],[A]&[H A] are molar concentrations of hydronium ion, conjugate base and weak acid at equilibrium. Ms. Bui is cognizant of metacognition and learning theories as she applies them to her lessons. Clearly, the pH of any solution must approach that of pure water as the solution becomes more dilute. First, let's write the balanced dissociation reaction of, Plugging the equilibrium concentrations into our. Taking the positive one, we have [H+] = .027 M; A 0.75 M solution of an acid HA has a pH of 1.6. However, make sure can do the 5%-thing for exams where Internet-accessible devices are not permitted! - Here we have a titration curve for the titration of 50 milliliters of 0.200 molar of acetic acid, and to our acetic solution we're adding some 0.0500 molar sodium hydroxide. The usual advice is to consider Ka values to be accurate to 5 percent at best, and even more uncertain when total ionic concentrations exceed 0.1 M. As a consequence of this uncertainty, there is generally little practical reason to express the results of a pH calculation to more than two significant digits. Plots of this kind are discussed in more detail in the next lesson in this set under the heading ionization fractions. (An exact numeric solution yields the roots 0.027016 and 0.037016). The approximation 0.10 x 0.10gives us, x (Ka Ca) = (0.010 0.10) = (.001) = .032(ii). Removal of a second proton from a molecule that already carries some negative charge is always expected to be less favorable energetically. And when, as occasionally happens, a quadratic is unavoidable, we will show you some relatively painless ways of dealing with it. a, b, and c, and away you go! This plot shows the combinations of Ka and Ca that generally yield satisfactory results with the approximation of Eq 4. The latter mixtures are known as buffer solutions and are extremely important in chemistry, physiology, industry and in the environment. Unless the solution is extremely dilute or. Thus [H+] = 101.6 = 0.025 M = [A]. With the exception of sulfuric acid (and some other seldom-encountered strong diprotic acids), most polyprotic acids have sufficiently small Ka1 values that their aqueous solutions contain significant concentrations of the free acid as well as of the various dissociated anions. Sorry, if it is. AP Chemistry Skills Practice. Is this a stupid question? The usual advice is that if this first approximation of x exceeds 5 percent of the value it is being subtracted from (0.10 in the present case), then the approximation is not justified. Acetic Acid (aka Ethanoic Acid) is the weak, monoprotic acid found in vinegar, and has a Ka= 1.76 x 10 -5 . However, it also exposes you to the danger that this approximation may not be justified. Cancel any time. x2 = 0.010 (0.10 x) = .0010 .01 x which we arrange into standard polynomial form: Entering the coefficients {1 .01 .001} into an online quad solver yields the roots Of those that do, the one at the MathIsFun site is highly recommended; others can be found here and at the Quad2Deg site. Calculate the Ka value of a 0.50 M aqueous solution of acetic acid ( CH3COOH ) with a pH of 2.52. Boric acid is sufficiently weak that we can use the approximation of Eq 1-22 to calculate a:= (5.8E10 / .1) = 7.5E-5; multiply by 100 to get .0075 % diss. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. x-term in the denominator. a) Calculate the pH of a 0.050 M solution of CO2 in water. Calculate the pH of a solution of a weak monoprotic weak acid or base, employing the "five-percent rule" to determine if the approximation 2-4 is justified. Example: Consider the process by which we would calculate the H 3 O +, OAc -, and HOAc concentrations at equilibrium in an 0.10 M solution of acetic acid in water. ], https://en.wikipedia.org/wiki/Acid_dissociation_constant#Acidity_in_nonaqueous_solutions, The base dissociation constant (or base ionization constant). If not, under what conditions would be higher (e.g. In order to predict the pH of this solution, we must first find [H+], that is, x. copyright 2003-2023 Study.com. But Ka for nitrous acid is a known constant of $$Ka \approx 1.34 \cdot 10^{-5} $$. Key points: For a generic monoprotic weak acid \text {HA} HA with conjugate base \text {A}^- A , the equilibrium constant has the form: This cycle is repeated until differences between successive answers become small enough to ignore. The percent dissociation for weak acid. It only takes a few minutes to setup and you can cancel any time. The pH of an aqueous acid solution is a measure of the concentration of free hydrogen (or hydronium) ions it contains: pH = -log [H +] or pH = -log [H 3 0 + ]. Diplomacy Overview, Types & Examples | What is Diplomacy? x (0.010 x .012) = (1.2E4) = 0.0011, Applying the "five percent rule", we find that x / Ca = .0011/.01 = .11 which is far over the allowable error, so we must proceed with the quadratic form. You don't have to use them, but it often is one of the best ways to keep track of lots of different numbers. The real roots of a polynomial equation can be found simply by plotting its value as a function of one of the variables it contains. for the example 1: calculating the % dissociation, the part where the ICE table is used and you can use the quadratic formula to find concentration "x", the two answers I got for x was x= -0.01285M and x=0.01245M. Find the Ka K a of a certain acid if 0.20 M solution of the acid has a pH of 3.00. . The dissociation fraction, \[ = \dfrac{[\ce{A^{}}]}{[\ce{HA}]} = \dfrac{0.025}{0.75} = 0.033\]. Working this out yields (1.5E4)/(.05) = .003, so we can avoid a quadratic. which is often expressed as a per cent (\(\alpha\) 100). A weak acid gives small amounts of H 3O + and A . 7. pH of an aqueous solution with acetic acid and sodium acetate. If you feel the need to memorize stuff you don't need, it is likely that you don't really understand the material and that should be a real worry! However, for almost all practical applications, one can make some approximations that simplify the math without detracting significantly from the accuracy of the results. \[ \ce{CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- } \nonumber\], According to the definition of pH (Equation \ref{eq1}), \[\begin{align*} -pH = \log[H_3O^+] &= -4.88 \\[4pt] [H_3O^+] &= 10^{-4.88} \\[4pt] &= 1.32 \times 10^{-5} \\[4pt] &= x \end{align*}\], According to the definition of \(K_a\) (Equation \ref{eq3}, \[\begin{align*} K_a &= \dfrac{[H_3O^+][CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]} \\[4pt] &= \dfrac{x^2}{0.2 - x} \\[4pt] &= \dfrac{(1.32 \times 10^{-5})^2}{0.2 - 1.32 \times 10^{-5}} \\[4pt] &= 8.69 \times 10^{-10} \end{align*}\]. Should I drop the x, or forge ahead with the quadratic form? Using the above approximation, we get This latter effect happens with virtually all salts of metals beyond Group I; it is especially noticeable for salts of transition ions such as hexaaquoiron(III) ("ferric ion"): This comes about because the positive field of the metal enhances the ability of H2O molecules in the hydration shell to lose protons. -2x or some other number)? Notice that the products of this reaction will tend to suppress the extent of the first two equilibria, reducing their importance even more than the relative values of the equilibrium constants would indicate. \[K_{\mathrm{a}} = \dfrac{x \cdot x}{c_{\mathrm{a}} - x}\]. If the acid is fairly concentrated (usually with Ca > 103 M) and sufficiently weak that most of it remains in its protonated form HA, then the concentration of H+ it produces may be sufficiently small that the expression for Ka reduces to. What is the Ka of the weak acid? Plug all concentrations into the equation for \(K_a\) and solve. As an example of how one might approach such a problem, consider a solution of ammonium formate, which contains the ions NH4+ and HCOO-. It will, of course, always be the case that the sum, For the general case of an acid HA, we can write a mass balance equation. General Chemistry:Principles & Modern Applications; Ninth Edition, Pearson/Prentice Hall; Upper Saddle River, New Jersey 07. I was trying to figure out which of the "x" is the correct one ( I assume since a negative concentration can not exist, the concentration has to be 0.01245M) and I have gone through my calculations a few times, and I don't know where I went wrong. First, let's write out the base ionization reaction for ammonia. Nevertheless, this situation arises very frequently in applications as diverse as physiological chemistry and geochemistry. The only commonly-encountered salts in which the proton is donated by the cation itself are those of the ammonium ion: \[\ce{NH_4^{+} NH)3(aq) + H^{+}\lable{2-6}\]. Solutions of glycine are distributed between the acidic-, zwitterion-, and basic species: Although the zwitterionic species is amphiprotic, it differs from a typical ampholyte such as HCO3 in that it is electrically neutral owing to the cancellation of the opposite electrical charges on the amino and carboxyl groups. is incomplete. When given the pH value of a solution, solving for \(K_a\) requires the following steps: Calculate the \(K_a\) value of a 0.2 M aqueous solution of propionic acid (\(\ce{CH3CH2CO2H}\)) with a pH of 4.88. Estimate the pH of a 0.10 M aqueous solution of HClO2, Ka = 0.010. What Is Osteogenic Sarcoma? What you've calculated using the quadratic formula is correct. the solution pH is log .027 = 1.6. This yields the positive root x = 0.0099 which turns out to be sufficiently close to the approximation that we could have retained it after all.. perhaps 5% is a bit too restrictive for 2-significant digit calculations! will be affected by the hydrogen ion concentration, and thus by the pH. You then substitute this into (2-2), which you solve to get a second approximation. {eq}Ka = \frac{\left [ H_{3}O^{+}\right ]\left [CH_{3}COO^{-} \right ]}{\left [ CH_{3}COOH \right ]} {/eq}, Step 4: Using the given pH, solve for the concentration of hydronium ions present with the formula: {eq}\left [ H_{3}O \right ]^{+} = 10^{-pH} {/eq}, {eq}\left [ H_{3}O \right ]^{+} = 10^{-2.52} {/eq}, {eq}\left [ H_{3}O \right ]^{+} = 0.003019 M {/eq}. A solution of CH3NH2 in water acts as a weak base. Rewriting the equilibrium expression in polynomial form gives, Inserting the coefficients {1 .022 .000012} into a quad-solver utility yields the roots 4.5E3 and 0.0027. A rigorous treatment of this system would require that we solve these equations simultaneously with the charge balance and the two mass balance equations. So for a solution made by combining 0.10 mol of pure formic acid HCOOH with sufficient water to make up a volume of 1.0 L, Ca = 0.10 M. However, we know that the concentration of the actual species [HCOOH] will be smaller the 0.10 M because some it ends up as the formate ion HCOO. Solution: From the stoichiometry of HCOONH4. If glycine is dissolved in water, charge balance requires that, \[H_2Gly^+ + [H^+] \rightleftharpoons [Gly^] + [OH^] \label{3-3}\], Substituting the equilibrium constant expressions (including that for the autoprotolysis of water) into the above relation yields. The difficulty, in this case, arises from the numerical value of Ka differing from the nominal concentration 0.10 M by only a factor of 10. Amines, a neutral nitrogen with three bonds to other atoms (usually a carbon or hydrogen), are common functional groups in organic weak bases. It expresses the simple fact that the "A" part of the acid must always be somewhere either attached to the hydrogen, or in the form of the hydrated anion A. Solve for the concentration of \(\ce{H3O^{+}}\) using the equation for pH: \[ [H_3O^+] = 10^{-pH} \]. Legal. 1. In most practical cases, we can make some simplifying approximations: In addition to the three equilibria listed above, a solution of a polyprotic acid in pure water is subject to the following two conditions: Material balance: although the distribution of species between the acid form H2A and its base forms HAB and A2 may vary, their sum (defined as the "total acid concentration" Ca is a constant for a particular solution: Charge balance: The solution may not possess a net electrical charge: Why do we multiply [A2] by 2? What about a salt of a weak acid and a weak base? This, of course, is a sure indication that this treatment is incomplete. Ka is represented as {eq}Ka = \frac{\left [ H_{3}O^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]} {/eq}. Compare the successive pKa values of sulfuric and oxalic acids (see their structures in the box, above right), and explain why they should be so different. However, for students in more advanced courses, this "comprehensive approach" (as it is often called) illustrates the important general methodology of dealing with more complex equilibrium problems. Unlock Skills Practice and Learning Content. Don't bother to memorize these equations! Step 2: Create an Initial Change Equilibrium (ICE) Table for the. With a Ka of 0.010, HClO2 is one of the "stronger" weak acids, thanks to the two oxygen atoms whose electronegativity withdraws some negative charge from the chlorine atom, making it easier for the hydrogen to depart as a proton. pKb = log \Kb = log (4.4 1010) = 3.36. Chemistry questions and answers. Estimate the pH of a 0.0100 M solution of ammonium formate in water. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Steps in Determining the Ka of a Weak Acid from pH Step 1: Write the balanced dissociation equation for the weak acid. What is interesting about this last example is that the pH of the solution is apparently independent of the concentration of the salt. When dealing with acid-base systems having very small Ka's, and/or solutions that are extremely dilute, it may be necessary to consider all the species present in the solution, including minor ones such as OH. We can treat weak acid solutions in much the same general way as we did for strong acids. However, who want's to bother with this stuff in order to solve typical chemistry problems? OAc - ( aq) Ka = 1.8 x 10 -5. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. It's important to understand that whereas Ka for a given acid is essentially a constant, \(\alpha\) will depend on the concentration of the acid. Solution: Because K1 > 1, we can assume that a solution of this acid will be completely dissociated into H3O+ and bisulfite ions HSO4. Step 5: Solving for the concentration of hydronium ions gives the x M in the ICE table. On right, structure of a generic amine: a neutral nitrogen atom with single bonds to R1, R2, and R3. This means the left side must be equally small, which requires that the denominator be fairly large, so we can probably get away with dropping x. AP European History: Renaissance Philosophy, Art & NY Regents Exam - Living Environment Help and Review Chapter 18: The Electromagnetic Spectrum and Light. Either method will yield the solution, Now that we know the concentration of hydroxide, we can calculate. This is an interesting area, but I don't know much about this I certainly don't remember learning about this in 1st or 2nd year undergraduate chemistry. A big \(K_a\) value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. Example: Given a 0.10M weak acid that ionizes ~1.5% [H +] = [A_] = 0.015(0.10)M = 0.0015M [H A] 0.10M 0.0015M 0.0985M You can find the percent ionization of a weak base. How to calculate weak acid concentration given pH and Ka Futurehope 327 subscribers Subscribe 89 Share 10K views 5 years ago A Level Chemistry - Paper 1 (Exams From 2018) In this. This energy is carried by the molecular units within the solution; dissociation of each HA unit produces two new particles which then go their own ways, thus spreading (or "diluting") the thermal energy more extensively and massively increasing the number of energetically-equivalent microscopic states, of which entropy is a measure. However, acid-base reactions definitely take place in solvents other than water and even in the gas phase. A diprotic acid H2A can donate its protons in two steps: In general, we can expect Ka2 for the "second ionization" to be smaller than Ka1 for the first step because it is more difficult to remove a proton from a negatively charged species. It's pretty straightfor. Substitute these values into equilibrium expression for \Kb: To make sure we can stop here, we note that (3.6E4 / .01) = .036; this is smaller than .05, so we pass the 5% rule and can use the approximation and drop the Remember, our goal was to calculate the pH of the resulting solution. H 2 O ( l) H 3 O + ( aq) +. . Examples of Magical Realism in Life of Pi. 2. In this article, we will discuss acid and base dissociation reactions and the related equilibrium constants: Weak acids are acids that don't completely dissociate in solution. Get unlimited access to over 88,000 lessons. If you are only armed with a simple calculator, then there is always the venerable quadratic formula that you may have learned about in high school, but if at all possible, you should avoid it: its direct use in the present context is somewhat laborious and susceptible to error. The value of [A -][H +] should be lower than the value of [HA] in order for Ka to be low. An unknown weak acid with a concentration of 0.088 M has a pH of 1.80. Direct link to tyersome's post This is an interesting ar, Posted 6 years ago. She has prior experience as an organic lab TA and water resource lab technician. pH of a polyprotic acid (LindaHanson, 17 min). Thus the only equilibrium we need to consider is the dissociation of a 0.010 M solution of bisulfite ions. The aluminum ion exists in water as hexaaquoaluminum Al(H2O)63+, whose pKa = 4.9, Ka = 104.9 = 1.3E5. 0.10 moles divided by 0.200 liters, gives the concentration of acetate anions of 0.50 molar. Calculate the pH at the equilibrium point in an acetic acid sodium hydroxide titration. In case of the strong acid and base we can directly use the concentration of the compound given because it dissociates totally. TExES English as a Second Language Supplemental (154) Advanced Excel Training: Help & Tutorials. Even if the acid or base itself is dilute, the presence of other "spectator" ions such as Na+ at concentrations much in excess of 0.001 M can introduce error. I am correct right? At left, structure of pyridine. Plus, get practice tests, quizzes, and personalized coaching to help you Direct link to yuki's post You can find the percent , Posted 6 years ago. OH- is actually considered to be a strong base, as its conjugate acid, water (H2O), is a weak acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which salts of the acid are present. The reason for this is that if b2 >> |4ac|, one of the roots will require the subtraction of two terms whose values are very close; this can lead to considerable error when carried out by software that has finite precision. The calculations shown in this section are all you need for the polyprotic acid problems encountered in most first-year college chemistry courses. As we will explain farther on, in most practical cases we can make some simplifying approximations which eliminate the need to solve a quadratic. For more on Zwitterions, see this Wikipedia article or this UK ChemGuide page. It can be used to calculate the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in an aqueous solution. Estimate the pH of a 0.20 M solution of acetic acid. For the more dilute acid, a similar calculation yields 7.6E4, or 0.76%. Solve for the concentration of H 3O + using the equation for pH: [H3O +] = 10 pH Use the concentration of H 3O + to solve for the concentrations of the other products and reactants. Two moles of H3O+ are needed in order to balance out the charge of 1 mole of A2. It is probably more satisfactory to avoid Le Chatelier-type arguments altogether, and regard the dilution law as an entropy effect, a consequence of the greater dispersal of thermal energy throughout the system. Strong acids and strong bases refer to species that completely dissociate to form ions in solution. Calculating pH when weak base is added to an strong acid. ICE tables are just a way of organizing data. If you're seeing this message, it means we're having trouble loading external resources on our website. This indicates that the . Ammonia will accept a proton from water to form ammonium, From this balanced equation, we can write an expression for, To determine the equilibrium concentrations, we use an, This is a quadratic equation that can be solved by using the quadratic formula or an approximation method. Salts of a strong base and a weak acid yield alkaline solutions. Get access to thousands of practice questions and explanations! Accessibility StatementFor more information contact us atinfo@libretexts.org. These acids are listed in the order of decreasing Ka1. Science. Salts such as sodium chloride that can be made by combining a strong acid (HCl) with a strong base (NaOH, KOH) have a neutral pH, but these are exceptions to the general rule that solutions of most salts are mildly acidic or alkaline. Because the successive equilibrium constants for most of the weak polyprotic acids we ordinarily deal with diminish by several orders of magnitude, we can usually get away with considering only the first ionization step. This approximation will not generally be valid when the acid is very weak or very dilute. Hold off rounding and significant figures until the end. which reminds us the "A" part of the acid must always be somewhere! This can be a great convenience because it avoids the need to solve a quadratic equation. For all acid-base equilibrium calculations that are properly set up, these roots will be real, and only one will be positive; this is the one you take as the answer. x = [H+] 1.9 103 M, and the pH will be log (1.9 103) = 2.7, b) Percent dissociation: 100% (1.9 103 M) / (0.20 M) = 0.95%. It is represented as {eq}pH = -Log[H_{3}O]^+ {/eq}, Become a member to unlock the rest of this instructional resource and thousands like it. Because the Ca term is in the denominator here, we see that the amount of HA that dissociates varies inversely with the concentration; as Ca approaches zero, [HA] approaches Ca. Research Methods and the Study of Human Growth and Vectors, Matrices & Determinants: Lesson Plans. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Setting x = [H+] = [Al(H2O)5OH 2+], the equilibrium expression is. Under certain conditions, these events can occur simultaneously, so that the resulting molecule becomes a double ion which goes by its German name Zwitterion. We therefore expand the equilibrium expression. FTCE General Knowledge Test (GK) (827): Reading Subtest OUP Oxford IB Math Studies: Online Textbook Help, UExcel Human Resource Management: Study Guide & Test Prep, Nick Carraway in the Great Gatsby: Character Analysis. However, don't panic! Find the pH of a 0.15 M solution of aluminum chloride. The usual approximation yields, However, on calculating x/Ca = .01 0.15 = .07, we find that this does not meet the "5% rule" for the validity of the approximation. Step 2: Using the definition. In the following development, we use the abbreviations H2Gly+ (glycinium), HGly (zwitterion), and Gly (glycinate) to denote the dissolved forms. The only difference is that we must now take into account the incomplete "dissociation"of the acid. Direct link to Dulyana Apoorva's post I guess you are correct, , Posted 3 years ago. Direct link to Bibika's post After reading the article, Posted 4 months ago. Figure 16.6.1: Some of the common strong acids and bases are listed here. He has over 20 years teaching experience from the military and various undergraduate programs. Although this is a strong acid, it is also diprotic, and in its second dissociation step it acts as a weak acid. Calculate the Ka value of a 0.021 M aqueous solution of nitrous acid( HNO2) with a pH of 3.28. This video shows how you can calculate the Ka of an acid, if you're given the pH of the solution (and its concentration, of course). The numerical value of \(K_a\) is used to predict the extent of acid dissociation. The protons can either come from the cation itself (as with the ammonium ion NH4+), or from waters of hydration that are attached to a metallic ion. ALEKS: Calculating the Ka of a weak acid from pH 6,219 views Apr 12, 2021 51 Dislike Share Save Roxi Hulet 7.53K subscribers How to use pH to calculate Ka. Determine the Ka K a of a certain . y = ax2 + bx + c, whose roots are the two values of x that correspond to y = 0. A 0.10 M solution of this amine in water is found to be 6.4% ionized. (The value of pKb is found by recalling that Ka + Kb = 14.). To calculate pH all you need is the H + ion concentration and a basic calculator, because it is a very straightforward calculation. in which Kb is the base constant of ammonia, Kw/109.3. Ah, this can get a bit tricky! and thus the acid is 3.3% dissociated at 0.75 M concentration. Let's go through this example step-by-step! Relating Ka and Kb to pH, and calculating percent dissociation. Relating Ka and Kb to pH, and calculating percent dissociation. Direct link to Katie Schleicher's post OH- is actually considere, Posted 6 years ago. Chemistry How to Calculate the. [I don't remember how to write equilibrium constant expressions], [Why is this called a "base dissociation constant" when the base doesn't dissociate? which, you will notice, as with the salt of a weak acid and a weak base discussed in the preceding subsection predicts that the pH is independent of the salt's concentration. However, without getting into a lot of complicated arithmetic, we can often go farther and estimate the additional quantity of H+ produced by the second ionization step. All explained in Section 3 of the next lesson. Measurement of electrically charged particles is a substance's pH. This important property has historically been known as hydrolysis a term still used by chemists. Solution: When methylamine "ionizes", it takes up a proton from water, forming the methylaminium ion: Let x = [CH3NH3+] = [OH] = .064 0.10 = 0.0064. Six Strong Acids. In sulfuric acid, the two protons come from OH groups connected to the same sulfur atom, so the negative charge that impedes loss of the second proton is more localized near the site of its removal. However, it will always be the case that the sum, If we represent the dissociation of a Ca M solution of a weak acid by, then its dissociation constant is given by. Drive Student Mastery. By contrast. Direct link to Ernest Zinck's post It's not a stupid questio, Posted 7 years ago. Petrucci,et al. According to the above equations, the equilibrium concentrations of A and H+ will be identical (as long as the acid is not so weak or dilute that we can neglect the small quantity of H+ contributed by the autoprotolysis of H2O).

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