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## Acids and bases

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**Acids and bases**Acids and bases have an important role in the context of water quality and water treatment processes. The composition and molar ratio between weak acids systems determines the pH value and the buffering capacity of the solution. The pH has a direct effect (biological activity, particle and species concentration etc.) and indirect effect (solid precipitation, gas dissolution, adsorption processes, flocculation etc.) over the quality of water.**Water ionization 2H2O H3O+ + OH-**Water is both an acid and a base = 10-14at 25 oC The activity of pure water is one (by definition); thus: Kw = 10-14= (H+)(aq)(OH-)(aq)**The p notation**pH = -log(H+) pOH = -log(OH-) pKa = -log(Ka) Kw = 10-14 = (H+) (OH-) log (10-14) = log (H+) + log (OH-) 14 = pH + pOH**Logarithm rules**Log(X*Y) = Log(X) + Log(Y) Log(X/Y) = log(X) – log(Y) Log10(X) = Y → 10Y = X pH = -log(H+) → 10-pH = (H+) Log(X2) = 2 log(X) Log(1) = 0**Acid base equilibrium**Strong acid or base – completely dissociate (!) at the standard pH range. Weak acid/base - materials that release or accept an hydrogen ion but undergo only partial ionization in the pH range of natural water. Equilibrium is maintained between the ionic species and non-ionic species, making for a buffering capacity. Mono protic weak base/acid (a weak base/acid system composed of two species that are distinguished by one proton): HA H+ + A- Basic species Acidic species**Clarification of the pK concept**Mathematically, pKais the negative logarithm value of the equilibrium constant. pKa=-log(Ka) In weak acid/base systems, the pK is also the pH value where the acidic species active concentration equals the basic species active concentration. pH = pKaAH = A- At this exact pH the weak acid system has the highest buffering capacity - in other words, the best ability to oppose change in pH as a result of dosage to the solution of a strong base or acid.**Derivation of speciation functions**Equilibrium equation AT= HA + A- Total A mass balance A little bit of algebra:**Examples of monoprotic weak acid/base systems**• Ammonia (pK=9.25)NH3 / NH4+ • Acetic acid/acetate (pK=4.76)CH3COOH / CH3COO- • Sulfide system (practical) (pK=6.99) H2S / HS- • Nitrite system (pK=3.70)HNO2/ NO2- • Hypochlorous acid (pK=7.60) HOCl / OCl-**pH curve: log(species) + buffering capacity curve of a**monoprotic weak acid with pK=6 HA A- OH- ( H+ ( Buffering capacity curve**AT = [H2A] + [HA-] + [A2-]**Sum of species concentrations: Diprotic systems H2A H+ + HA- HA- H+ + A2- Each species can be isolated as a function of pH, ATand equilibrium constants Example:**The expression**The Species The system Monoprotic Bi-protic Tri-protic**pK1**pK2 H2A HA- A2- AT log (species) H+ OH-**In Recirculated Aquaculture System (RAS) NH3<1 mg/l as N is**permitted. Given total ammoniacal nitrogen (TAN, NH3-N) of 400 mg/l as N and pH = 7. What is the concentration of NH3? What is the limiting pH to attain NH3 < 1 mg/l? What is the CO2 How much acid needed to obtain the needed pH? (later in the course…….)**Dosing acids to pure water**HCl / NaOH AH / NaA**Proton Balance Equation (PBE)**Each H+donated by a molecule does not “disappear”, rather accepted by another molecule. For example, when water and nitrite acid release protons: Basic state: H2O HNO2 H3O+ PBE +H+ [H+] = [NO2-] + [OH-] -H+ -H+ NO2- OH- • Analytical process for solving simple acid/base equilibrium problems • Write and balance the relevant chemical equations including water ionization. • Write the expressions of the equilibrium constants • Write a species mass balance equation for all the participating systems • Write a balance equation (for charges or proton transfer) • Solve for all variables Or… Write PBE Substitute the expressions from the table in the last slide Solve for the unknowns**Example: strong acid**Calculate the pH of an HCl solution at 10-6 M (assume negligible ionic strength and 25 oC) Solution: HCl H+ + Cl- H2O H+ + OH- ‘ Chemical equations Mass balance ClT= [HCl] + [Cl-] = 10-6 M [H+] = [Cl-] + [OH-] Charge balance (or proton balance) After changing sides we get: [H+] = 10-6M pH =6**Approximated solution**For strong acids we can assume that the reaction continues all the way: HCl H+ + Cl- Additionally, we can assume that the water ionization equation is not significant: the concentration of H+ ions created by water ionization is negligible compared to the H+ concentration released by the acid. According to the above assumptions, the H+ concentration equals the added HCl concentration 10-6 M pH = -log (H+) = -log (10-6) = 6.0**Example 2: weak acid**Calculate the pH of nitric acid at a concentration of 10-3 M (assume negligible ionic strength and 25 oC): HNO2 H+ + NO2- (1) (pKa = 3.70) (2) NT = [HNO2] + [NO2-] Mass balance (3) (4) [H+] = [NO2-] + [OH-] (PBE) Proton balance equation: We get: (same equation, of course, only with different constants)**H+**+H+ -H+ -H+ NO2- OH- [H+] = [NO2-] + [OH-] Proton balance Every H+ ion released by one species does not disappear but rather is absorbed by a different species: H2O Initial state: HNO2 Calculation of Ka from Gibbs energy tables Solve for [H+]by knowing NT = 10-3 M G0reaction = [(-8.25 + (0)] – [(-13.3)] = 5.05 Kcal Ka = 10- G0/1.363 = 1.97 • 10-4 Analytical solution pH = 3.45**Approximate solution**Water ionization is neglected (!) Define X = the fraction of nitric that has undergone ionization in equilibrium: [NO2-] = X [H+] = X [HNO2] = 0.001 – X A cubic equation is solved: This type of solution is suitable for simple systems only X =3.56 • 10-4 M pH = -log X = 3.45 At very low or high pK values one should take caution when neglecting water ionization**Graphical solution of base/acid equilibrium problems based**on a pH- log (species) graph In the past this was the preferred way to solve complex systems. The resulting solution is an approximation and (usually) also accurate enough for the needs of water treatment processes. The graphical method is good enough for attaining adequate pH results which is a result of mixing a number of weak acids.**Monoprotic strong acid (HCl)**• Write the relevant chemical equations: • HCl H+ + Cl- H2O H+ + OH- • 2. Write the equilibrium constants equations:K’w = (H+)[OH-] • 3. Write the mass balanceClT = [Cl-] + [HCl] • Define the species for which you intend to draw curves (in this case H+, OH-, Cl- ) • 5. Arrange all the values as logs: • log [Cl-] = log CT, log [H+] = -pH, log[OH-] = pH – pKw • All the equations are linear on a logarithmic graph, for the slope check the derivative: = [Cl-]**New pH**Cl- For solving, define the proton balance: [H+] = [Cl-] + [OH-] At the intersection point of Cl- and OH-, the common line increases by 0.3 log units above the line describing the chloride concentration, because: log{[Cl-] + [OH-]} = log{2[Cl-]} = log 2 + log[Cl-] = log [Cl-] + 0.3**Graphical expression of a monoprotic weak acid system**(example: the ammonia system) • NH4+ H+ + NH3 • H2O H+ + OH- • NT = [NH4+] + [NH3] • Relevant species: OH- ,H+ ,NH3 ,NH4+ • The species equations on a logarithmic basis: • For the water system, as before log [NH4+] = log NT- pH - log([H+] + K’a) This equation is clearly non-linear with respect to pH**Simplifying assumptions (only algebra)**Case 1: [H+]>>KaorpKa>pH When [H+] is greater than Ka by at least one order of magnitude, Ka can be neglected and the equation becomes: log [NH4+] = log NT – pH –log([H+])= log NT Since the slope is nil, the concentration curve at a distance greater than 1 pH from the pK is a horizontal straight line at a concentration of NT Case 2 : [H+]<<Ka or pKa< pH When [H+] is smaller than Ka by at least one order H+ can be neglected and the equation becomes: log [NH4+] = log CT – pH – log Ka The graph slope:**The same is done for NH3, yielding:**Note: The graph is a general case for a given HA/A-**Drawing the pK area**At pH = pK , the two species concentrations are equal, therefore: [NH3] = [NH4+] = 0.5 NT log [NH3] = log [NH4+] = log(NT/2) = log NT – log 2 = log NT – 0.3**Solving problems using the graphical approach**Example: Nitric acid HNO2(pKa=3.7) was added to distilled water (pH=7) at 10-2 M. What will be the pH of the solution? Solution: The final pH will be reached when all the weak acid/base systems in the water reach equilibrium. In this case there are only two systems: the nitric and the water systems. The missing equation is the proton mass balance Proton balance [H+] = [NO2-] + [OH-]**New pH (HNO2EP)**[H+] = [NO2-] + [OH-]**Equivalence points**The pH value reached by distilled water which was mixed with a weak acid species is defined as an equivalence point of the solution, and the solution is defined as an “equivalent solution” • Importance of equivalence points • Is used as a reference point for defining alkalinity and acidity values and for titration-end points required for determining alkalinity and acidity values • In many cases the equivalent point is the pH where the buffering ability is minimal • The equivalent point is the pH value to which the solution often converges to when the species is added. For example, when bicarbonate (HCO3-) is added, the pH value will move towards the equivalent point of bicarbonate (pH 8.3) irrespective of the initial pH.**Proton balance:**HAc H+ H2O Ac- HNO2 NO2- OH- Another example for using graphical solutions: Nitrous acid (HNO2)pK=3.7 at 10-3 M, and sodium acetate (NaCH3COO) 10-2 M (pKAc=4.75 (was added to distilled water. What is the resulting pH? Solution Nitrous acidHNO2 ↔ H+ + NO2- Sodium acetate is a strong electrolyte that dissociates completely to Na+ and acetate ion (CH3COO-) The acetate itself is a weak base (Ac-/Hac) HAc ↔H+ + Ac- Initial state: [HAc] + [H+] = [NO2-] + [OH-]**New pH**[HAc] + [H+] = [NO2-] + [OH-]**Graphical expression of a diprotic system**AT = [H2A] + [HA-] + [A2-] Mass balance [H2A], [HA-], [A2-], [H+], [OH-] Participating species Logarithmic equation calculations **A logarithm is applied on both equation sides:**log[H2A] = log AT + 2log [H+] – log ([H+]2 + [H+]K’a1+K’a1K’a2) log[H2A] = log AT – 2pH - log ([H+]2 + [H+]K’a1+K’a1K’a2) pKa2>pKa1>pHor alternativelyCase 1: Ka2<<Ka1<<[H+] Any expression containing Ka1 or Ka2 can be neglected: log[H2A] = log AT – 2pH - log [H+]2 = log AT Curve slope Conclusion At low pH values that are lower by at least 1 pH unit than pKa1, the curve of [H2A] as a function of pH is horizontal and equal to AT**K’a2<<[H+]orCase 2 : pK’a2>pH**Every term that contains K’a2 in the expression ([H+]2 + [H+]K’a1+K’a1K’a2) can be neglected. This results in: log[H2A] = log AT – 2pH - log ([H+]2 + [H+]K’a1) This equation represents a curve and it can be applied at range of 1 pH from both sides of pKa1 Case 3 :K’a2<<[H+]<<K’a1 or alternatively pK’a2>pH>pK’a1 The terms [H+]2 ו K’a1K’a2 can be neglected, yielding: log[H2A] = log AT – 2pH - log [H+] + log K’a1= log AT – pH + pK’a1 The slope:**[H+]<<K’a1orCase 4 : pH>pK’a1**[H+]2 can be neglected, therefore: log[H2A] = log AT – 2pH - log ([H+]K’a1+K’a1K’a2) This results in a nonlinear curve which can be applied at range of 1 pH from both sides of pKa2 K’a2[H+]<<orCase 5: pH>pK’a2 [H+]Ka1and [H+]2 can be neglected. log[H2A] = log AT – 2pH – (log K’a1+ logK’a2) = log AT – 2pH + pK’a1 + pK’a2 The slope:**The equations for [HA-]and[A2-] can be calculated in the**same way (something to do at home) Drawing curves in the vicinity of pK’a1 and pK’a2 : When: pH = pK’a1: [H2A] = [HA-] and [A2-] = 0 Therefore: log [H2A] = log AT – 0.3 log [HA-] = log AT – 0.3 When pH = pK’a2 [A2-] = [HA-] and [H2A] = 0 Therefore: log [A2-] = log AT – 0.3 log [HA-] = log AT – 0.3**H+**Effect of the AT value on the curves AT does not affect the general shape of the curves apart for their location on the Y axis (will go upwards for higher AT and downwards for lower AT) AT**Tri-protic systems**Example: Phosphate system H3PO4, H2PO4-, HPO42-, PO43- pKp1=2.12 pKp2=7.2 pKp3=12.7 Equation development – during the exercise