Acids and Bases Topic # 14 Brønsted-Lowry Definitions

Acids and Bases Topic # 14 Brønsted-Lowry Definitions
The Ion Product for Water The pH and Other “p” Scales Aqueous Solutions of Acids and Bases Hydrolysis The Common Ion Effect Buffer Solutions Indicators and Titrations Polyprotic Acids

Arrhenius definitions
Acid Anything that produces hydrogen ions in a water solution. HCl (aq) H+ + Cl- Base Anything that produces hydroxide ions in a water solution. NaOH (aq) Na OH- Arrhenius definitions are limited to aqueous solutions.

Brønsted-Lowry definitions
Expands the Arrhenius definitions Acid Proton donor Base Proton acceptor This definition explains how substances like ammonia can act as bases. NH3(g) + H2O(l) NH OH-

Strong acids and bases considered to ionize completely. HCl(aq) + H2O(l) H3O+ (aq) + Cl-(aq) NaOH(aq) + H2O(l) Na+(aq) + OH-(aq) Weak acids and bases do not ionize completely. HC2H3O2 (aq) + H2O(l) H3O+(aq) +C2H3O2-(aq) NH3 (aq) + H2O(l) NH4+ (aq) + OH- (aq)

When a Brønsted-Lowry acid dissolves in water, the water acts as a base. HC2H3O2 (aq) + 2H2O(l) H3O+(aq) + C2H3O2-(aq) Acid Base Acid Base H3O+ (aq) is called the hydronium ion. However, the exact number of water molecules required to hydrate a proton is not known.

Conjugate acid-base pairs. Acids and bases that are related by loss or gain of H+ as H3O+ and H2O. Examples. Acid Base H3O+ H2O HC2H3O2 C2H3O2- NH4+ NH3 H2SO4 HSO4- HSO4- SO42-

Common acids and bases Acids Formula Molarity* nitric HNO3 16
hydrochloric HCl sulfuric H2SO4 18 acetic HC2H3O2 18 Bases ammonia NH3(aq) 15 sodium hydroxide NaOH solid *undiluted.

Common Acids and bases Acidic Basic Citrus fruits Baking soda
Aspirin Detergents Coca Cola Ammonia cleaners Vinegar Tums and Rolaids Vitamin C Soap

HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
Water is an amphoteric substance that can act either as an acid or a base, HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) acid base acid base H2O(l) + NH3(aq) NH4+(aq) + OH-(aq) acid base acid base

Auto-ionization of water
Auto-ionization When water molecules react with each another to form ions. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) (10-7M) (10-7M) Kw = [ H3O+ ] [ OH- ] = 1.0 x at 25oC Note: [H2O] is not included in expression because H2O is a pure liquid! ion product of water

Ion product for water Kw is a temperature dependent equilibrium constant. We commonly use 25oC as the standard temperature. Temperature, oC Kw x 10-15 x 10-15 x 10-14 x 10-14 x 10-14

Autoionization of water
[H+] and [OH-] are always present in aqueous solutions. Only for a neutral solution are they at the same concentration. Neutral solution. [H+] = 10-7 M = [OH-] Acidic solution. [H+] > 10-7 M > [OH-] Basic solution. [H+] < 10-7 M < [OH-]

pH and other “p” scales We need to measure and use acids and bases over a very large concentration range. And many times knowing those concentrations is very important. pH and pOH are systems to keep track of these very large ranges. pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14

pH calculations Determine the following. pH = -log[H+]
pH of 6.7 x 10-3 M H+ = 2.2 pH of 5.2 x M H+ = 11.3 [H+], if the pH is 4.5 = 3.2 x 10-5 M H+

pOH examples Determine the following. pOH = -log[OH-] = 14 - pH
pOH of 1.7 x 10-4 M NaOH pOH = pH = 10.2 pOH of 5.2 x M H+ pH = pOH = 2.7 [OH-] , if the pH is 4.5 pOH = 9.5 [OH-] = 3.2 x M

pH scale A logarithmic scale used to keep track of the large changes in [H+]. 10-14 M M M Very Neutral Very Basic Acidic When you add an acid, the pH gets smaller. When you add a base, the pH gets larger.

pH of some common materials
Substance pH 1 M HCl Gastric juices Lemon juice Classic Coke Coffee Pure Water Blood Milk of Magnesia Household ammonia 1M NaOH

Acid and Base Strength Strong acids Ionize completely in water. HCl, HBr, HI, HClO3, HNO3, HClO4, H2SO4. Weak acids Partially ionize in water. Most acids are weak. Strong bases Ionize completely in water Strong bases are metal hydroxides - NaOH, KOH Weak bases Partially ionize in water. no Ka ! 100% Ionization! no Kb ! 100% Ionization!

These bases are considered to be strong.
LiOH lithium hydroxide NaOH sodium hydroxide KOH potassium hydroxide RbOH rubidium hydroxide CsOH cesium hydroxide *Ca(OH)2 calcium hydroxide *Sr(OH)2 strontium hydroxide *Ba(OH)2 barium hydroxide * Completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100%. The other five in the list can easily make solutions of 1.0 M and are 100% dissociated at that concentration.

Acid and Base Strength For strong acids and bases, we can directly calculate the pH or pOH if we know the molar concentration. Examples. 0.15 M HCl would produce 0.15 M H+. pH = -log(0.15) = 0.82 pOH = pH = = 0.052 M NaOH produces M OH-. pOH = -log(0.052) = 1.28 pH = pOH = =

Acid dissociation constant, Ka
The ionization of a weak acid can be expressed as an equilibrium. HA (aq) + H2O(l) H3O+(aq) + A- (aq) The strength of a weak acid is related to its equilibrium constant, Ka. Ka = We omit water. It’s a pure liquid. [A-][H3O+] [HA]

HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)
Weak acid equilibria Example Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = x 10-5 HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq) The first step is to write the equilibrium expression. Ka = [H3O+][Bz-] [HBz]

Weak acid equilibria HBz H3O+ Bz- Initial conc., M 0.10 0.00 0.00
Change, DM x x x Eq. Conc., M x x x [H3O+] = [Bz-] = x We’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.

Weak acid equilibria Solve the equilibrium equation in terms of x
Ka = x = x = (6.28 x )(0.10) = M pH = 2.60 x2 0.10

Dissociation of bases, Kb
The ionization of a weak base can also be expressed as an equilibrium. B (aq) + H2O(l) BH+(aq) +OH- (aq) The strength of a weak base is related to its equilibrium constant, Kb. Kb = [OH-][BH+] [B]

Ka and Kb values For weak acids and bases
Ka and Kb always have values that are smaller than one. Acids with a larger Ka are stronger than ones with a smaller Ka. Bases with a larger Kb are stronger than ones with a smaller Kb. For a conjugate acid:base pair, pKa + pKb = 14 = pKw Most acids and bases are weak.

Ionization constants at 25oC.
Acid pKa pKb Acetic acid Ammonium ion Benzoic acid Formic acid Lactic acid Phenol

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
Hydrolysis Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction. CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) This type of reaction is given a special name. Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH-. The reaction of a cation with water to produce the conjugate base and H3O+.

Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq)
Hydrolysis There is only a small # of cations with a high enough positive charge that can act as acids in hydrolysis. Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq) These metal ions are able to pull electrons from the H-OH bond. If the pull is strong enough, water is split. One of the surrounding water molecules will take on the H+ and form H3O+.

Common ion effect (Just an example of Le Châtelier’s principle.)
The shift in equilibrium caused by the addition of an ion formed from the solute. Common ion An ion that is produced by more than one solute in an equilibrium system. Adding the salt of a weak acid to a solution of weak acid is an example of this.

Common ion effect Example
What is the concentration of hydrogen ions in a solution formed by adding mol of sodium acetate to one liter of M acetic acid (HAc). (Ac = C2H3O2-) (assume that the volume of the solution does not change when the salt is added) [Ac-] [H+] [HAc] Ka = = 1.7 x 10-5

Common ion effect HAc Ac- H+ Init. Conc., M Change, DM -x +x +x Eq. Conc., M x x x To solve for x, substitute the equilibrium values into the expression. Ka [HAc] [Ac-] (1.7 x 10-5) (0.099-x) ( x) x = =

Common ion effect If we assume that x is negligible compared to and 0.097, we can simplify the problem to: x = x = 1.7 x 10-5 = [H+] Note. When a solution contains equal concentrations of an acid and its conjugate base, [H+] = Ka. (1.7 x 10-5) (0.099) (0.097)

Ions of Neutral Salts Salt Hydrolysis
A salt is formed between the reaction of an acid and a base. HCl + NaOH  NaCl + H2O Usually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction. CATIONS Na+ K+ Rd+ Cs+ Mg+2 Ca+2 Sr+2 Ba+2 ANIONS Cl- Br- I- ClO4- BrO4- ClO3- NO3-

These ions have little tendency to react with water, they just
In this NEUTRAL environment, all the ions are spectator ions from the strong acid and strong base reaction. There are no insoluble particles. These ions have little tendency to react with water, they just stay dissociated into 100% ions. So, salts consisting of these ions are called neutral salts and the resulting solution has a pH of 7. For example: NaCl, KNO3, CaBr2, CsClO4 are neutral salts.

When weak acids and weak bases react, the strength of the stronger ion conjugate acid or conjugate base in the salt determines the pH of its solutions. Acidic Ions NH4+ Al+3 Pb+2 Sn+2 Transition Metal Ions HSO4- H2PO4- Basic Ions F- C2H3O2- NO2- HCO3- CN- CO3 -2 S-2 SO4-2 HPO4-2 PO4-3

The resulting salt solution can be acidic, neutral or basic.
A salt formed between a strong acid and a weak base is an acid salt, for example NH4Cl: ( HCl + NH3  NH4Cl ) Strong weak The NH4Cl is soluble and reacts with H2O: NH H2O  NH3 + H3O+ The act of the salt particles reacting with water molecules in the ‘salt solution’ is HYDROLYSIS.

A salt formed between a weak acid and a strong base is a basic salt, for example NaCH3COO.
(CH3COOH + NaOH  NaCH3COO + H2O) CH3COO H2O  CH3COOH + OH- pH > 7

Deciding whether a salt is acidic, basic or neutral!
******Decide which ion in the salt could produce a stronger acid or stronger base. Example: NaF NaOH or HF NaOH is the stronger base so NaF is a basic salt NaClO NaOH or HClO4 BOTH are strong so neutral salt! Fe(NO3) Fe(OH)2 or HNO3 HNO3 is the stronger compound Fe(NO3)2 is an acidic salt

What if both ions are from weak acids and weak bases?
Salt: NH4NO NH4OH and HNO2 weak base and weak acid Can compare Kb for NH4OH and Ka for HNO2: Kb NH4OH = 1.6 x10-5 Ka for HNO2 = 4.5 x 10-4 Ka > Kb so the salt is acidic!

Ka2 for H2CO3 = 4.8 x 10 -11 So (NH4)2CO3 is basic!
Another example: ( qt.1h. on Salt Hydrolysis WS) (NH4)2CO NH4OH and HCO3- weak base and weak acid ion Kb for NH4OH = 1.6 x 10 -5 Ka2 for H2CO3 = x So (NH4)2CO3 is basic!

[HCN] [OH-] [CN-] Calculations !!! Kb = Kw Ka 1 x10-14 5.8 x 10-10
Calculate the pH of a M solution of KCN. ( Ka for HCN is 5.8 x 10-10) Salt is a basic salt, so need to use Kb…. CN- + H2O  HCN + OH- 0.5M x x - ion makes OH- ! [HCN] [OH-] [CN-] Kb = Kw Ka 1 x10-14 5.8 x 10-10 Kb = 1.7 x10-5 = =

x2 0.5 –x 1.7 x10-5 = / x = 2.9 x 10-3 M pOH = -log(2.9 x 10-3) = = = pH

HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Buffers Solutions that resist pH change when small amounts of acid or base are added. Two types weak acid and its salt weak base and its salt HA(aq) + H2O(l) H3O+(aq) + A-(aq) Add OH Add H3O+ shift to right shift to left Based on the common ion effect. Attacks HA

Buffers solution with this organization: HC2H3O2  H+ + C2H3O2-
A solution of a weak acid and its’ salt! Acetic Acid and Sodium Acetate: HC2H3O NaC2H3O2 Mix these two compounds in water gives a solution with this organization: HC2H3O2  H C2H3O2- Concentration of the C2H3O2- would be greater than just a solution of low soluble HC2H3O2 !!!!

How does a buffer work? Ethanoic acid and ethanoate:
Both: CH3CO2 - AND CH3CO2H are present in solution CH3CO H3O+  HCH3CO2 + H2O The ethanoate ions will be able to absorb any excess acid that is added HCH3CO2 + OH-   CH3CO H2O The ethanoic acid will be able to absorb any excess base that is added

Buffers Henderson-Hasselbalch equation The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two. Calculating pH of an acidic buffer use: pH = pKa + log Calculating the pH of a basic buffer use: pH = 14 - ( pKb + log ) [A-] [HA] [HA] [A-]

Buffers The beauty of a buffer….
A total of 100 ml of a 1.0 M HCl is add in 10 ml increments to 100 ml of each of the following: Solution A: Pure water - pH = 7 (no acid in it ! ) Solution B: A solution containing 1.0 M HA (acid in it ), 1.0 M A- (conjugate base in it ) with a pKa of 7.0 Calculate the pH after each addition.

Buffers Initially, each sample is at pH 7.00
After adding 10 ml of 1.0 M HCl we have: Pure water [H3O+] = = pH = 1.04 From pH 7 to pH 1.04….. This is a pretty big jump! (10 ml)(1.0 M) (110 ml)

Buffers CH3CO2- + H3O+  HCH3CO2 + H2O
Add 10 ml of 1.0 M HCl to our buffered system. .10 moles moles moles - .01 moles moles moles We started with 0.10 moles of both the acid and conjugate base forms. The HCl in our first 10 ml can be expected to react with the conjugate base, converting it to the acid. After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form. CH3CO H3O+  HCH3CO2 + H2O

Buffers Checking the pH:
Since all we need to be concerned about is the ratio of A- to HA in the 10-90% region then: pH = pKA + log = log = 6.91 A change of only 0.09 (Buffer solution) compared to 5.96 ( in pure water) ! [A-] [HA] 0.09 mol 0.11 mol

Buffers Matter of fact…….. ml HCl pH added unbuffered buffered

Buffers buffered unbuffered pH ml HCl added

Indicators Acid-base indicators are highly colored weak acids or bases. HIndic Indic- color color 2 They may have more than one color transition. Example. Thymol blue Red Yellow Blue One of the forms may be colorless - phenolphthalein (colorless to pink)

Indicators Indicator color change.
Indicators are commonly used to detect the endpoint of an acid-base titration. The indicator should not start to undergo a color change until about one pH unit after the equivalence point.

Indicators Selection of an indicator.
Under these conditions, the transition range for our indicator is pKa + 1. Ideally then, you want your indicator pKa to be one above the pKa for an acid sample. For bases, it should be one lower. You can seldom find a ‘perfect’ indicator so should use one that is a maximum of one unit away.

Indicators pH transition Indicator range color
Bromophenol Blue yellow-blue Methyl Orange red-yellow Methyl Red red-yellow Bromothymol Blue yellow-blue Cresol Purple yellow-purple Phenolphthalein colorless - pink Thymolphthaleine colorless - blue Alizerin Yellow GG yellow - red

Indicator examples Acid-base indicators are weak acids that undergo a color change at a known pH. pH phenolphthalein

Indicator examples bromthymol blue methyl red

Titrations revisited Methods based on measurement of volume.
If the concentration of an acid is known, the amount of a base can be found. If we know the concentration of the base, then we can determine the amount of acid. All that is needed is some calibrated glassware and either an indicator or pH meter.

Titrations Buret - volumetric glassware used for titrations.
It allows you to add a known amount of your titrant to the solution you are testing. If a pH meter is used, the equivalence point can be measured. An indicator will give you the endpoint.

Titrations Note the color change which indicates that the ‘endpoint’ has been reached. Start End

Titration curves Acid-base titration curve
A plot of the pH against the amount of acid or base added during a titration. Plots of this type are useful for visualizing a titration. It also can be used to show where an indicator undergoes its color change.

Titration curves Equivalence pH Point % titration or ml titrant
Four regions of titration curve Overtitration Indicator Transition Equivalence Point pH Buffer region % titration or ml titrant

Titration curves Strong acid titrated with a strong base.
This is pretty straight forward since the net reaction is: H3O+ + OH H2O Prior to equivalence point The pH indicates the amount of sample present after accounting for dilution. [H3O+] = moles acid - moles base total volume in liters

Equivalence point [H3O+] = [OH-] pKW = 14 = pH + pOH pH = 7.00 So the equivalence point for strong acid/base problems is always at pH=7.00

Overtitration Past the equivalence point, we don’t have any acid remaining. All that we are doing is diluting our titrant by adding more solution volume and more OH- ions. [OH-] = pH = 14 - pOH moles excess total volume in liters

Titration curves Example.
Construct a titration curve for the titration of 100 ml of 0.10 M HCl with 0.10 M NaOH A. 0% titration pH = -log[0.10] = 1 B. After 10 ml NaOH [H3O+] = = M, pH = 1.09 (100 ml)(0.10 M) - (10 ml)(0.10 M) 100 ml + 10 ml

Titration curves ml titrant total ml [H3O+] pH 0 100 0.10 1.00

Titration curves pH ml NaOH Let’s add 10ml more!

Titration curves pH ml NaOH

Titration curves Reached the Equivalence point or 100 % titration
(100ml acid)(0.10M acid) = (100ml base)( 0.10M base) pH MUST be x10-14 = (1 x10-7)2 Note that for the first 90 ml of our titration, we only saw a change of 1.28 pH units. Now we have a jump of 4.72 pH units.

Titration curves Anything over the equilvalence point is
Overtitration: (no more acid there!) Describing the dilution of our titrant: 10 ml overtitration [OH-] = M = M pOH = 2.32 pH = = (.10M)10 ml 210 ml

Titration curves ml titrant total volume [OH-] pH 110 210 0.0048 11.68
Continue the calculations: ml titrant total volume [OH-] pH

Titration curves Just adding more OH- Over titration pH ml NaOH

Titration curves Titration of a strong base with a strong acid.
This is not significantly different from our earlier example. If you plot pOH rather that pH, the results would look identical. Typically we still plot pH versus ml titrant so the curves are inverted.

Titration curves basic sample (base in the beaker) pH
acidic sample (acid in the beaker) ml titrant

Titration of weak acids or bases
Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example. We must be concerned with conjugate acid/base pairs and their equilibria. Example HA + H2O H3O+ + A- acid base

First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid. We still have the same four general regions for our titration curve. The calculation will require that you use the appropriate Ka or Kb relationship. We’ll start by reviewing the type of calculations involved and then work through an example.

If your sample is an acid then use Ka = At this point [H3O+] = [A-]. You can solve for [H3O+] by using either the quadratic or approximation approach. Finally, calculate the pH. [H3O+][A-] [HA]

If your sample is an base then use Kb = At this point [OH-] = [HA]. You can solve for [OH-] by using either the quadratic or approximation approach. Then determine pH as 14 - pOH. [OH-][HA] [A-]

In this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our answer. A common format for the equilibrium expression used in this region is the Henderson-Hasselbalch equation. We can present it in two forms depending on the type of material we started with.

Starting with an acid pH = pKa + log Starting with a base pH = 14 - ( pKb + log ) ( We’re just determining the pOH and then converting it to pH. ) [A-] [HA] [HA] [A-]

Another approach that can be taken is to simply use the % titration values. For an acidic sample, you would use: pH = pKa + log % titration 100 - % titration

These equations have their limits and may break down if: Ka or Kb > 10-3 You are working with very dilute solutions. In those cases, you need to consider the equilibrium for water

100% titration - the equivalence point. At the point, we have converted all of our sample to the conjugate form. If your sample was an acid, now solve for the pH using the Kb relationship -- do the opposite if you started with a base. Remember that pKa + pKb = 14 You must also account for dilution of your sample as a result of adding titrant.

Overtitration (> 100%) These calculations are identical to those covered in our strong acid/strong base example. You simply need to account for the amount of excess titrant added and how much it has been diluted.

Example A 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve. For benzoic acid Ka = x 10-5 pKa = 4.20

Before adding any base: Ka = [H3O+] = [A-] [HA] + [A-] = 0.10 M (We’ve assumed that [A-] is negligible compared to [HA].) [H3O+][A-] [HA]

Initial pH, before adding any base: Ka = x = x = (6.28 x )(0.10) = M pH = 2.60 x2 0.10

adding 10ml increments of base: Here we can use the Henderson-Hasselbalch equation in ml titration format pH = pKa + log pH = log (10 / 90) = 3.25 We can calculate other points by repeating this process. (increments of 10ml) (100ml – total added) (Showing the first increment of 10ml)

ml added pH Note: At 50 ml titration, pH = pKa Also, the was only a change of 1.91 pH units as we went from 10 to 90 ml titration. (5.15 – 3.24)

pH % titration

100% titration (EQUILVALENCE POINT) At this point, virtually all of our acid has been converted to the conjugate base - benzoate. Need to use the Kb relationship to solve for this point! (have no acid left!) Kb = Kb = Kw / Ka = x 10-10 [OH-] [HA] [A-]

EQUILVALENCE POINT At the equivalence point: [HA] = [OH-] (We’ve diluted the [A-] = 0.05 M sample and the total volume at this point is 200 ml) Finally, assume that [benzoic acid] is negligible compared to [benzoate]. .10m .200 liters

100% titration ( EQUILVALENCE POINT) Kb = x = x = (1.59 x )(0.050) = x 10-6 M pOH = pH = 14 - pOH = 8.45 x2 0.050

pH % titration

Overtitration All we need to do here is to account for the dilution of our titrant. 10 % overtitration (10 ml excess) [OH-] = M = M pOH = 2.32 pH = =

ml titrant total volume [OH-] pH This is identical to what we obtained for our strong acid/strong base example

Titration of weak acids
pH % titration

Polyprotic acids A number of acids exist that have more that one ionizable hydrogen. Examples H3PO4 Phosphoric Acid H2SO4 Sulfuric Acid H2C2O4 Oxalic acid H2CO3 Carbonic acid Each H+ will ionize with its own constant. Also, it becomes more difficult to remove subsequent H+.

Stepwise equilibrium Example - H3PO4
Each hydrogen is capable of dissociation but not to the same extent at any given pH. H3PO H2PO HPO PO43- Since the removal of any one H+ affects how easily subsequent H+ are removed, each step has its own Ka value. Ka1 Ka2 Ka3

Stepwise equilibrium Note: [H3O+] is the same for each expression.
The relative amounts of each species can be found if the pH is known. The actual amounts can be found if pH and total H3PO4 is known. [H3O+] [H2PO4-] [H3PO4] KA1 = [H3O+] [HPO42-] [H2PO4-] KA2 = [H3O+] [PO43-] [HPO42-] KA3 =

Titration Curve for Diprotic Acid

Acids and Bases Topic # 14 Brønsted-Lowry Definitions

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Acids and Bases Topic # 14 Brønsted-Lowry Definitions

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