Acids, Bases, pH

Bronsted-Lowry Definition

A Brønsted-Lowry acid is a proton (hydrogen ion) donor. A Brønsted-Lowry base is a proton (hydrogen ion) acceptor.

Neutralization

A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water. Consider the reaction between HCl and NaOH in water: HCl(aq)acid+NaOH(aq)base ⇋ NaCl(aq)salt+H₂O(l) This can be written in terms of the ions: H⁺ + Cl⁻ + Na⁺ + OH⁻ → Na⁺ + Cl⁻ + H⁺ + OH⁻ When the spectator ions are cancelled, we get just H₂O. Neutralization is the basis of titration. A pH indicator shows the equivalence point—the point at which the equivalent number of moles of a base have been added to an acid. It is often wrongly assumed that neutralization should result in a solution with pH 7.0; this is only the case in a strong acid and strong base titration.

Arrhenius Definition

An Arrhenius acid dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H⁺ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions, OH⁻.

Strong or Weak Acid/Base?

Strong acids/bases completely dissociate into H⁺/OH⁻. Weak acids/bases do not. Whether or not an acid is likely to give up its hydrogen proton depends on the stability of its conjugate base, which can then be measured with other factors: electronegativity and atomic radius. These two factors also affect bond strength. A strong acid has weak bond strength so that it's easy to give up its hydrogen proton to get to its conjugate base (if it's more stable). A more electronegative element is more stable when they have a negative charge resting on them, which explains why polarity increases with acidity. In general the more s-character the bond is, the closer the electron is to the nucleus. (an electron is negatively charged while a nucleus is positively hence the closer the electron to the nucleus the more stable it is). So if the bond with the acidic hydrogen is sp instead of sp3, it will be more acidic since sp has 50% s-character as opposed to 25%. If the conjugate base of the molecule is very stable, which it is in this case (the conjugate base has the electron all by itself), then the molecule is acidic. When these compounds act as an acid, an H-X bond is broken to form H+ and X- ions. The more polar this bond, the easier it is to form these ions. Another reason to explain this is because if a molecule is more polar, it's be easier to be ionized by the also polar water molecules. This is why the more oxygen the acid has, the stronger the acid. Thus, the more polar the bond, the stronger the acid. Another factor is atomic radius. Charges prefer to be on larger atoms than on smaller atoms. This preference results from large atoms allowing the negative charge to delocalize over a much larger region of space, instead of being concentrated in a small region (as it would on a small atom). As a rule, atom size trumps electronegativity considerations. So, even though fluorine is a more electronegative atom than iodine, HI is more acidic than HF. The much larger iodine atom (which would be the conjugate base) allows the negative charge to delocalize over a larger space than does the much smaller fluorine atom, and thus makes hydrogen iodine more acidic. We can also think of it this way: the bigger the atomic radius, the longer the bond length (bigger atoms have more shells, and bond length is defined as the average distance between two atomic nuclei). The longer the bond length, the weaker the bond strength which allows the hydrogen proton to be stripped away more easily. Bond strength also plays a role. The stronger the bond, the less acidic it is. This also explains why HF is weak since its a strong bond due to low radius (F is all the way to the right).

Temperature and pH

Temperature plays a significant role on pH measurements. As the temperature rises, molecular vibrations increase which results in the ability of water to ionise and form more hydrogen ions. As a result, the pH will drop. Every solution will undergo a change in their pH value through changes in temperature. A difference in pH measurements at different temperatures is NOT an error! The new pH level simply tells about the true pH for that solution at that specific temperature.

Ka and pKa/Conjugate pairs

The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid in solution. Ka and Kb refers to the conjugate acid and base respectively, so that means that when you multiply them together, you will get Kw. This makes sense if you write it out mathematically (Ka∙Kb). What this tells us is that if Ka is very large, Kb must be very small and vice verca. This means that the stronger the acid, the weaker the conjugate base. If you add a proton, you get the conjugate acid. You remove the proton and you get the base. If it's a strong conjugate acid, then the dissociation is strongly favorable, which means its not favorable to get back that proton. An example would be HCl--it would completely dissociate into H+ and Cl-. Since its very favorable, it wouldn't be favorable for Cl- to get back that proton. This is another way of explaining why a strong conjugate acid must have a weak conjugate base and vice versa. Unlike pH, Ka is more specific in that it tells you how strong or weak the acid is. When you dissolve an acid in water, it undergoes an equilibrium reaction with the water: HA + H₂O ⇌ H₃O⁺ + A⁻ where HA is a generic acid that dissociates into A−, known as the conjugate base of the acid and a hydrogen ion which combines with a water molecule to make a hydronium ion. The value of the equilibrium constant is given by: Ka = [H₃O⁺][A⁻]/[HA] We ignore water because liquids and solids are excluded from the equilibrium constant. For polyprotic acids (acids that give more than one H+), there are two Ka values. The first Ka value is for the first proton it donates, while the second Ka value is for the second proton. An example is H2SO4, sulfuric acid. The first Ka is 1.2x10⁻² for the equilibrium expression [H+][HSO3-]/[H2SO3]. The second Ka is 6.4x10⁻⁸ for [H+][SO3²⁻]/[HSO3-]. Note how the expression changes since you remove the H+ twice for polyprotic acids. The second Ka is also a lot smaller than the first Ka since it's harder to remove a proton when the molecule already lost one, ie. it's positively charged (same logic for I.E. energy). Ka essentially tells us what the ratio of the products to reactants is at equilibrium, which gives us an idea which way does the equation favor more and thus how frequent the acid dissociates (the strength). So we can conclude the greater the Ka, the stronger the acid. Ka will give us an exponential number, so we convert these exponential numbers into a normal range by taking their negative logarithm just like the pH. So, pKa=-log₁₀Ka. The smaller the value of pKa, the stronger the acid. We only use Ka and pKa for acids. If we want to measure the strength of the conjugate base, we must use Kb and pKb.

Why pKa=pH at Half Equivalence Point

pH = pKa + log [A⁻]/[HA] At the half equivalence point, say we have 10 moles of WA. So, there will be 5 moles of SB as the name suggests (half equivalence). So the 10 moles of WA will be neutralized to 5 moles by the SB. When this happens, the CB will also equal to the number of WA which is 5. You can visualize this by drawing a CBA table. There was 0 CB before and the change was -5 for WA and SB (they neutralize) and +5 for the CB of the WA. And because this is occurring at the half equivalence point, the number of CB will always be equal to the WA after neutralization. [WA]=[CB] Thus, using the HH equation, pH=pKa since the log term equals to zero.

pOH

As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH = -log₁₀[OH⁻] [OH⁻]=10⁻ᵖᴴ The pOH scale is similar to the pH scale in that a pOH of 7 is indicative of a neutral solution. A basic solution has a pOH of less than 7, while an acidic solution has a pOH of greater than 7. The pOH is convenient to use when finding the hydroxide ion concentration from a solution with a known pH. Remember, pOH only refers to the concentration of OH⁻, not H⁺. However, we can figure out the pH of based on the fact that pH + pOH = 14 at 25 C.

Buffer Capacity and HH Equation

Buffer capacity is the amount of acid or base with which a buffer can react before giving a significant pH change. It increases if you increase the concentration of the weak conjugate acid-base pair. This makes sense since the more acid-base, the more it can absorb the strong acid/base. Therefore, we can conclude that if the concentration of a strong acid/base is greater than that of the conjugate pair, then the buffer will no longer be effective. Furthermore, buffers are usually prepared with equal concentrations of conjugate acid-pair so that the buffer will have equal capacity to neutralize either added base/acid, ie. If you're skewed too far one way or the other, then you will not have much of the respective species around to neutralize added acid or base. Note that this calls for equal HA and A- concentrations, and it's not a constant at equilibrium. [A-]/[HA] is not the same as [HA]/[A-][H+], which is the Kb of A-. Buffers work best when pKa=pH because the closest pKa is to the pH, then the closest [A-]/[HA] is to 1, as it should be for maximum capacity. This can be mathematically proven in the Henderson Hasselbalch equation, which relates pH, pKa, [HA], and [A-]: Ka=[H₃O⁺][A⁻]/[HA] [H₃O⁺]=Ka[HA]/[A⁻] -log[H₃O⁺]=-log Ka - log [HA]/[A⁻] (log properties) -log[H₃O⁺]=-log Ka + log [A⁻]/[HA] pH = pKa + log [A⁻]/[HA] As you can see, if pKa is equal to pH, then [A-]/[HA] is equal to 1, ie. The two concentrations are equal which is our goal. Likewise, if you set [A-]/[HA] equal to 1 and solve for pKa instead, you'll find that pKa will be equal to pH. Either way, we learn that if pKa is equal to or very close to pH, then the two concentrations of the acid-base pair will be equal or very close, making it an effective buffer. An alternative to using concentrations of the acid and base is using moles. It still conceptually makes sense because in either case, the goal is to have the ratio of base to acid be as close to 1 as possible. The ratio would be the same since if you use molarities, the liters cancel out in the denominators which leaves you just moles anyway. So, you can measure the quantities of the conjugate acid-base pair in both molarity and moles.

Weak Acid/Strong Base titration Calculation 1 (more acid than base)

Find the pH after adding 10 mL of 0.3 M NaOH in a titration of 25 mL of 0.3 M HF. The Ka value is 6.6×10−4 First thing we need to do is recognize that since this a titration involving a strong base, the reaction will proceed completely regardless of the strength of the acid. What this means is that we can subtract the moles of NaOH from whatever moles of HF we have to find the remaining number of HF. We have 0.01 L x 0.3M = 0.003 mol NaOH (or OH-) 0.025 L x 0.3M = 0.0075 mol HF We have more acid than base, so there should be acid remaining after 0.003 mol NaOH neutralizes with 0.003 mol HF. We can better visualize this by constructing a BCA table: HF + NaOH → H20 + F- B 0.0075 0.003 | 0 C -0.003 -0.003 | +0.003 A 0.0045 0 | 0.003 If you don't construct a BCA table, make sure to remember that when you react a WA with SB, you will always make the conjugate base, or F- in this case. After all the acid has neutralized the base, what we have left over is the acid and its conjugate base. This is when we can use the HH equation to calculate the pH since we essentially have a buffer solution. pH = pKa + log [A⁻]/[HA] But first, we need to calculate the new concentrations of acid and CB. We added 10mL of base to 25mL of acid, so new volume is 35mL or 0.035L. We also know pKa since Ka is given. [HF]=0.0045/0.035=0.1287 M [F-]=0.003/0.035=0.0857 M pKa=−log(6.6×10−4)=3.18 pH=−log(6.6×10−4)+log[.0857/0.1287] =3.00

Weak Acid/Strong Base titration Calculation 2 (equivalence point)

Find the pH after adding 25 mL of 0.3 M NaOH in a titration of 25 mL of 0.3 M HF. The Ka value is 6.6×10⁻⁴ First find moles of HF: 0.025L x 0.3M=0.0075 mol HF Then find moles of NaOH: 0.025L x 0.3M=0.0075 mol NaOH, or OH- Construct a BCA table: HF + NaOH → H20 + F- B 0.0075 0.0075 | 0 C -0.0075 -0.0075 | +0.0075 A 0 0 | 0.0075 This is the equivalence point of the titration. We know this because the acid and base are both neutralized and neither is in excess. To find the concentrations we must divide by the total volume. This is the initial volume of HF, 25 mL, and the addition of NaOH, 25 mL. Therefore the total volume is 25 mL + 25 mL = 50 mL or 0.05 L. [F-]=0.0075/0.05=0.15M However, to get the pH at this point we must realize that F-, a base, will hydrolyze. An ICE table for this reaction must be constructed: F- + H20 → HF + OH- I 0.15 | 0 0 C -x | +x +x E 0.15-x | x x Since F- is a base, we need the Kb value to find pH. Ka is given, so Kb would be 1.0x10⁻¹⁴/6.6×10⁻⁴=1.515x10⁻¹¹ The Kb equation would be x²/0.15-x=1.515x10⁻¹¹. Since Kb is extremely small, x is neglibile so this simplifies into x²/0.15=1.515x10⁻¹¹. x=1.5x10⁻⁶ pOH=5.82 pH=14 - 5.82=8.18

Percent Ionization

Percent ionization of an acid is the percent of acid dissociated. In a weak acid, the percent would be very low since its not favorable. In a strong acid, it would be very high. For weak acids, there is a trend of increasing percent ionization if you decrease the molarity of the weak acid. This can be explained by Le Chatelier's principle. The equilibrium expression for the weak acid is HA + H₂O⇌H₃O⁺ + A⁻. If your concentration is decreasing, that means the solution is being diluted, so the equilibrium is shifting to the left to counteract the increasing water. In other words, more acids dissociate when you dilute the solution according to Le Chatelier's principle.

Weak Acid/Strong Base titration Calculation 3 (more base than acid)

Find the pH after adding 26 mL of 0.3 M NaOH in a titration of 25 mL of 0.3 M HF. The Ka value is 6.6×10−4 First find moles of HF: 0.025L x 0.3M=0.0075 mol HF Then find moles of NaOH: 0.026L x 0.3M=0.0078 mol NaOH, or OH- Notice how we have more base than acid this time. Like in the first calculation, the base and acid will neutralize each other. Construct a BCA table: HF + NaOH → H20 + F- B 0.0075 0.0078 | 0 C -0.0075 -0.0075 +0.0075 A 0 0.0003 0.0075 Because we have more base than acid, there will be base remaining after neutralization. We have no more HF, 0.0003 mol OH-, and 0.0075 mol F-. How do we find the pH? Your first instinct might be to construct an ICE table after realizing that there are 0.0075 moles of F-. While it is true that these 0.0075 moles of F- will react with water to make more OH-, it is a negligible amount since the Kb of F- (1.0x10⁻¹⁴/6.6x10⁻⁴=1.5x10⁻¹¹) is so small. So while you could construct an ICE table and calculate the concentration of OH- after hydrolyzing F-, you would get a very similar number if you just took the -log of the current OH- concentration you have now. But before you do, you have to find the new concentration. The new volume is 26+25=51mL or 0.051L. The OH- concentration will therefore be 0.0003/0.051=0.006M OH-. -log(0.006)=2.23=pOH 14-2.23=11.77

Equilibrium Constant of Water

It is already known that water is able to self ionize into H₃O and H⁻ in very small amounts. We can calculate exactly how small their concentrations are by using the chemical equilibrium constant, K. It is essentially a ratio of the molarities of products (H₃O and H⁻) to reactants (H₂O) at the point of equilibrium. This equation excludes liquids and solids, so the concentration of water in this example is 1. H₃O and H⁻ are kept because they are aqueous. So, the equation is [H₃O][H⁻]=K which would give you 10⁻¹⁴, since the concentration of H₃O (10⁻⁷) multiplied by concentration of H⁻ (10⁻⁷) gives you 10⁻¹⁴. Remember, these values are from the fact that liter of water self ionizes into 10⁻⁷ ions each. They are in equal amounts since water is both an acid and a base. Of course, both ions neutralize themselves completely which gives you a neutral pH. Remember, this is only true at 25 C since temperature heavily affects pH. Since we already know that the concentration of both ions to be 10⁻⁷, we can take the -log₁₀ of that number to get our pH value, which is of course 7. Both concentrations of the ions in pure water are 7, so 7+7=14 which gives us our scale.

Indicators and color change when pH=pKa

The color change of a pH indicator is caused by the dissociation of the H+ ion from the indicator itself. Recall that pH indicators are not only natural dyes but also weak acids. The dissociation of the weak acid indicator causes the solution to change color. The equation for the dissociation of the H+ ion of the pH indicator is show below: HIn + H2O ⇌ H3O+ + In- where HIn is the acidic pH indicator and In- is the conjugate base of the pH indicator. If we take phenolphthalein as an example, the color of HIn would be colorless while the color of In- would be red. Phenolphthalein has a pKa of about 9.3, so it's a very weak acid and remains undissociated (this means it remains colorless at equilibrium). However, if you remove H+ by adding a base, then the equilibrium shifts to the right which produces In-, giving it a red color in the case of phenolphthalein. But how much base do you need to add in order to shift the equilibrium enough to the left in order to give it a red color? To answer this question, we apply the Ka equation: Ka= [H+][In-]/[HIn]. Indicators have a different color when in the molecular form as opposed to when in the ionic form, so it seems logical to assume that the point of changing color will be there is 50% of both forms present. In other words, the equilibrium equation lies exactly in the middle. If this happens, then [In-]=[HIn] in which case we could just cancel them out from the Ka equation to give the new equation Ka=[H+]. Therefore, pKa=pH if you take the -log of both sides. What does this mean? This means that the indicator changes color when pKa=pH. So the answer to the previous question is that you need to add enough base so that the pH becomes equal to the pKa of the indicator in order for it to change color.

Reaction of a Strong Acid and Base

The first step in writing an acid-base reaction is determining whether the acid and base involved are strong or weak as this will determine how the calculations are carried out. For reactions with strong acid and strong base, the net ionic equation will always be the same since the acid and base completely dissociate and the resulting salt also dissociates. This leaves the final product to simply be water. This is displayed in the following example involving hydrochloric acid (HCl) and sodium hydroxide (NaOH): HCl(aq)+NaOH(aq)→H₂O(l)+NaCl(aq) Since HCl and NaOH fully dissociate into their ion components, along with sodium chloride (NaCl), we can rewrite the equation as: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻(aq) We can simplify this equation by writing the net ionic equation of this reaction by eliminating the reactants that don't react (they dissociate in the solution. In this case, Na⁺ and Cl⁻). These reactants are known as spectator ions: H⁺(aq) + OH⁻(aq) → H₂O(l) The above equation describes the most important concept of a strong acid/strong base reaction, which is that a strong acid provides H⁺ ions (more specifically hydronium ion H₃O⁻) that combine with OH⁻ ions from a strong base to form water. One thing to note is that the anion of our acid HCl was Cl⁻(aq), which combined with the cation of our base NaOH, Na⁺(aq). This formed the salt NaCl(aq), which isn't shown in the net ionic equation since it dissociates and doesn't react with water. It is important, however, to remember that a strong acid/strong base reaction does form a salt, but that it is neutral (some salts are acidic/basic). The net ionic equation for a strong acid-strong base reaction is always: H+(aq)+OH⁻(aq)→H₂O(l) which is neutral.

Acidic, Basic, or Neutral Salt?

To determine whether a salt is acidic, basic, or neutral, we put them in aqueous solution. For example, we can see that the neutralization of NaCl is from a strong acid (HCl) and a strong base (NaOH). A strong acid and strong base always gives a pH of 7, so NaCl is a neutral salt. We can also figure this out by looking at NaCl's constituents: Na⁺ and Cl⁻, both of which cannot be hydrolyzed by water since both atoms has already filled its octets. You might think that Cl⁻ would react with water to form back HCl, but this won't happen since HCl is a very strong acid and so the chloride anion and water reaction won't be favorable. Therefore, NaCl is neutral. Likewise, an acidic salt would be one whose neutralization is from a strong acid and weak base, while a basic salt would have a neutralization from a strong base and weak acid. This is intuitive, since the strong base trumps the weak acid and vice versa. Another way would be looking at the dissociation of the salt. Consider HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(l). NaC₂H₃O₂, the salt product, can be dissociated into individual Na⁺ and C₂H₃O₂⁻ ions. We already know Na⁺ cannot be hydrolyzed by water. C₂H₃O₂⁻, however, can hydrolyze to form a basic solution: C₂H₃O₂⁻ + H₂O(l) ⇌ HC₂H₃O₂(aq) + OH⁻. This OH⁻ is what makes C₂H₃O₂⁻ basic, and thus NaC₂H₃O₂ a basic salt.

ICE and BCA tables to solve for pH

We use what's called an ICE table when we are solving titration problems that involve either a weak acid or weak base. When we do problems like these, we can't simply use MV=MV because that would imply that a weak acid/weak base ionizes completely, which isn't the case. That's what Ka and Kb are for--these equilibrium constants tell us to what extent do these weak acids/weak bases ionize. So if we can use these Ka and Kb values (depending on what we're solving for), we can mathematically determine the amount of H+ or OH- ions that's in solution when the titration is all done. To set up the expression, we use an ICE table, which stands for Initial, Change, and Equilibrium. Example: 70 mL of 1.25M HCN is titrated with 0.25M NaOH. Calculate the pH after the addition of (a) 0 mL of NaOH and (b) 100 mL of NaOH. The Ka of HCN is 6x10⁻¹⁰. For part a, we aren't adding any base. That means the only thing that's in solution is our weak acid, HCN, that's reacting with water. But remember that it's weak and so it doesn't dissociate completely. So we must set up an ICE table: HCN + H20 ⇌ H30⁺ + CN⁻ I 1.25M 0 0 C -x +x +x E 1.25-x x x (First of all, notice how I'm not calculating the number of moles of HCN and instead just using 1.25, the molarity. That's because we will use Ka later on, and Ka is an equilibrium constant that only uses concentrations and not moles. So that's why we only put concentrations in ICE tables.) Initially, there's no products yet. So the reaction must go to towards the right. Since the stoichiometric ratio is 1:1, for every x moles of HCN reacts with water, there are x moles of H30+ and CN-. The equilibrium result is that there are 1.25-x M of HCN and x M of H30+. This is when we use Ka, which is products over reactants. In this case, that is x/1.25-x. We set that equal to 6x10⁻¹⁰ and solve for x. You would get a quadratic equation, but we can avert that by making a simplifying assumption. We know that x (the H+ concentration) is going to be extremely small just by looking at HCN's Ka, so we can say that x is negligible and so 1.25-x will still be very close to 1.25. If we make that assumption, the equation becomes a lot easier to solve and x will be about 2.75x10⁻⁵. Remember that x is the concentration of H+, so we take the -log of this number and the pH will be about 4.56. It is acidic which is reasonable since we only had HCN in solution. To answer part b, we do NOT have to use ICE tables. Why? Because NaOH is a strong base, and whenever you react a strong base with a weak acid or vice versa the weak acid dissociates completely due to the presence of the strong base. This means that we CAN use MV=MV. To solve this problem, we use what's called a BCA table. BCA tables stand for Before, Change, and After, and they don't use Ka or Kb values so we don't use molarities. Rather, we use moles. HCN + NaOH → H20 + NaCN B 0.08750.025 0 C -0.025 -0.025 +0.025 A 0.0625 0 0.025 In the before row, there are 0.07x1.25M=0.0875 mol HCN and 0.1x0.25M=0.025 mol NaOH. There are 0 mol of NaCN since there aren't any products yet. Because NaOH is a strong base, all of HCN will react with NaOH. So on the change row, there are now 0.0875-0.025 mol HCN and 0.025-0.025 mol NaOH. There are now 0.025 mol HCN since the mole ratio is 1:1. To get the pH, we can use the Henderson-Hasselbach equation, pH=pKa+log[A-]/[HA]. pKa is the-log of 6x10⁻¹⁰, the Ka that's given. The moles of A-, the conjugate base (NaCN) is 0.025 and the moles of HA, the acid (HCN) is 0.0625. Notice how you can also use moles in place of molarity for the second part of the equation. This is because if you use molarity, the units would be in (mol/L)/(mol/L), which is the same thing as mol/mol. In other words, you can use both because the ratio would be the same regardless. The pH is 8.82 which is reasonable since the pH goes up after adding a strong base. It's also reasonable since it's lower than the pKa, which was 9.222, since there were more of the conjugate base than acid.

Buffer System

When there are too many H⁺ ions, a buffer will absorb some of them, bringing pH back up; and when there are too few, a buffer will donate some of its own H⁺ ions to reduce the pH. Buffers typically consist of an acid-base pair, with the acid and base differing by the presence or absence of a proton. If a solution is too acidic, the equation will move to the right so that the bicarbonate ion will absorb some of that H⁺. Likewise, the equation will move to the left so that carbonic acid will dissociate into H⁺. A buffer is simply a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffers work by reacting with any added acid or base to control the pH. For example, let's consider the action of a buffer composed of the weak base ammonia, NH3, and its conjugate acid, NH4+. When HCl is added to that buffer, the NH3 "soaks up" the acid's protons to become NH4+. Because the protons are locked up in the ammonium ion, the protons do not serve to significantly increase the pH of the solution. When NaOH is added to the same buffer, the ammonium ion donates a proton to the base to become ammonia and water. Here the buffer also serves to neutralize the base. As the above example shows, a buffer works by replacing a strong acid or base with a weak one. The strong acid's proton is replaced by ammonium ion, a weak acid. The strong base OH- was replaced by the weak base ammonia. These replacements of strong acids and bases for weaker ones give buffers their extraordinary ability to moderate pH.

pH

a measure of the hydrogen ion (H⁺) concentration of a solution. Below 7 is acidic, and above is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8. To calculate pH, we must know the concentration (molarity) of the substance. If that substance is acidic, we are able to find out how many H⁺ ions (or hydronium, since H⁺ reacts with water) there are in the solution. For example, if we know the concentration of H+ (acid) is .1 M, then we take the negative log₁₀ of .1, which would be a pH of 1. If the concentration is .01, then the pH would be 2. This makes sense because we already stated that each pH value is 10 times more acidic than the next value (10 times more H+ ions), which is why we're taking log₁₀. However, this simply taking the -log of a concentration won't always work. For example, the pH of a 10⁻⁹ M HCl won't be 9 because 9 would be basic. In this case, the concentration is so low that it's even below the autoionization of water. This means that we must take that into account by writing out the mathematical expressions. We can also convert pH to concentration, using the inverse equation: [H⁺]=10⁻ᵖᴴ. It's important to note that one can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar (one molar=pH of 0). Remember, pH only refers to the concentration of H⁺, not OH⁻. However, we can figure out the pOH based on the fact that pH + pOH = 14 at 25 C. This is always true in aqueous solutions. For example, orange juice is a mixture in water that is acidic. So, that acid will dissociate in aqueous solution and the hydrogen protons will react with some of the 10-7 hydroxide molecules in one liter of aqueous solution as well as water molecules to form hydronium. Since this lowers OH- and increases H+ concentration, the orange juice will have a pH less than 7, and the statement pH+pOH=14 is always true in aqueous solution at 25 C as H+ went up while OH- went down in the mixture of water and juice. pH is NOT a measure of the strength of the acid; it only measures how concentrated the acid is. To measure strength, we use pKa.

Self-ionization of Water

an ionization reaction in pure water or in an aqueous solution, in which a water molecule, H₂O, deprotonates (loses the nucleus of one of its hydrogen atoms) to become a hydroxide ion, OH⁻. This occurs because water is polar, so they can take away each other's hydrogen atoms. This is a very low possibility, however. In fact, only 10⁻⁷ molecules will self-ionize in one liter of water. Increasing the temperature will increase the chances of water ionizing, which is why pH is different at different temperatures. The hydrogen nucleus, H⁺, immediately protonates another water molecule to form hydronium, H₃O⁺. This is why taking the pH of H⁺ or H₃O⁺ is the same thing. Taking the -log of a concentration won't always work. For example, the pH of a 10⁻⁹ M HCl won't be 9 because 9 would be basic. In this case, the concentration is so low that it's even below the autoionization of water. This means that we must take that into account by writing out the mathematical expressions.