1. Section 14.1 Bronsted Acids and Bases
1. Acids - molecules that can lose H⁺ (proton donors) making H₃O⁺ in water (acids lose H⁺) 
2. Bases - molecules than can gain H⁺ (proton acceptors) and/or make OH^- in water (bases gain H⁺ / make OH^-)
3. Conjugate acid/base pairs - differ by one H⁺
1. when an acid loses one H⁺ it becomes its conjugate base
2. when a base gains one H⁺ it becomes its conjugate acid
3. examples (acid first then conjugate base) H₃PO₄ and H₂PO₄^-,  NH₄⁺ and NH₃, H₃O⁺ and H₂O, H₂PO₄^- and HPO₄^2-, H₂O and OH^-, CH₃COOH and CH₃COO^-
4. Notice that after losing H⁺ the charge goes down by one
4. Acid and Base reactions in water (proton transfer to or from water)
1. H₂CO₃(aq) + H₂O(l)  D  HCO₃^-(aq) + H₃O⁺(aq).   Label them as acid, base, conjugate acid (ca) ad conjugate base (cb) Answer:  H₂CO₃(aq) is the acid (carbonic acid),  H₂O(l) acts as the base here,   HCO₃^-(aq) is the cb,  H₃O⁺(aq) is the ca
2. CH₃NH₂ (aq) + H₂O(l)  D  CH₃NH₃⁺ (aq) + OH^-(aq).  Label them as acid, base, conjugate acid (ca) ad conjugate base (cb) Answer:  CH₃NH₂ (aq) is the base (an amine),  H₂O(l) acts as the acid here,   CH₃NH₃⁺ (aq) is the ca, OH^-(aq) is the cb
3. Remember that H⁺ is just short hand for H₃O⁺called hydronium ion.  H⁺ doesn't really exist in water.
4. Remember that water is not an acid or base, it can just act like one depending on the other molecules around it. 
5. Try example 14.1 - 2, problems 14.1 -3
2. Section 14.2 Acid and Base Strength
1. Predicting a direction to reach equilibrium.  You need to refer to a Table that has acids and conjugate bases in order of strength like Table 14.1 in McMurry Fay.  The stronger the acid, the more power it has to react, the more H⁺ it will give.  The stronger acid will react MORE than the weaker acid so the equilibrium will lie on the weaker acid side. 
1. Consider this reaction:  CH₃COOH + SO₄^2- D  CH₃COO^- + HSO₄^-.  There are two acids here:  acetic acid CH₃COOH and HSO₄^- ion.  Which is stronger???  HSO₄^- is stronger and so it will react more, and it reacts towards the left side (reactants).  So the equilibrium will lie on the reactant side - few products will be made.  We will mostly have acetic acid and sulfate ion in the beaker with just a bit of acetate ion and hydrogen sulfate ions in the beaker. K is less than one, K is a small number, reactants are favored.
2. Consider this reaction:  NH₃ + HF  D  NH₄⁺ + F^-.  The two acids are HF and NH₄⁺ ion.  Which is stronger?  HF is stronger so it will react more and it reacts towards the products.  The equilibrium lies on the product side. K is greater than one, products are favored. 
3. Try example 14.3 and problems 14.4-5.
2. Strong acids lose almost all of the H⁺ and ionize making hydronium ions, H₃O⁺ in water.  We say they ionize 100% and their reaction with water goes one way - no equilibrium is set up.  
1. Memorize the six strong acids:  hydrochloric acid HCl, hydrobromic acid HBr, hydroiodic acid HI, nitric acid HNO₃, perchloric acid HClO₄ and sulfuric acid, H₂SO₄.
2. All of them follow this pattern in water:  HCl (aq) + H₂O (l) g  Cl^- (aq) + H₃O⁺ (aq) which can also be written as just HCl (aq)  g  Cl^- (aq) + H⁺ (aq)
3. If I begin with 2.5M strong acid, I will have 2.5M hydronium ion and 2.5M conjugate base. 
4. A picture of HCl in water would have no HCl molecules, it would have only the ions since it all ionized. 
3. Weak acids lose just a small percentage of H⁺ and only partially ionize in water making hydronium ions. They set up equilibrium.
1. All other acids are weak except the 6 strong ones.  Four you should be familiar with are acetic acid CH₃COOH, phosphoric acid H₃PO₄, carbonic acid H₂CO₃ and hydrofluoric acid HF.
2. All weak acids follow this pattern in water:  HF (aq) + H₂O (l) D  F^- (aq) + H₃O⁺ (aq) which can also be written as just HF (aq)  D  F^- (aq) + H⁺ (aq)
3. If I begin with 2.5M weak acid, I do not know how much hydronium ion and conjugate base I have without calculating and ICE table.
4. A picture of HF in water would have mostly HF molecules with few ions.
4. Strong bases almost completely ionize in water making OH^-, hydroxide ions.   
1. Memorize the eight strong bases:  sodium hydroxide NaOH, lithium hydroxide LiOH, potassium hydroxide KOH, rubidium hydroxide RbOH, cesium hydroxide CsOH, calcium hydroxideCa(OH)₂, strontium hydroxide Sr(OH)₂ , and barium hydroxide Ba(OH)₂
2. They are all soluble solids and dissolve and dissociate almost completely in water just like:  NaOH (s)  g Na⁺(aq) + OH^-(aq)
3. If I begin with 2.0M NaOH I will get 2.0M hydroxide ion and 2.0M sodium ion.
4. A picture of KOH in water would have no KOH solid left, only K⁺ and OH^- ions.
5. Weak bases partially ionize in water making hydroxide ions. They set up an equilibrium.
1. The other metal hydroxides are largely insoluble thus in water they only dissolve and dissociate to a small extent.  Ammonia and other amines are weak bases because they react WITH water making hydroxide.
2. Other metal hydroxides like Mg(OH)₂ (s)  D Mg⁺²(aq) + 2 OH^-(aq)
3. Amines react with water like:  NH₃ + H₂O  D  NH₄⁺ + OH^-
4. If I begin with 1.0M ammonia, I don't know how much hydroxide I have without calculating.
5. A picture of ammonia in water has mostly NH₃ molecules with some ammonium and hydroxide ions.
6. Notes
1. A strong acid's conjugate base is really not basic but neutral (ie Cl^- from HCl is a neutral ion)
2. The stronger the acid, the weaker its conjugate base.  The stronger the base, the weaker its conjugate acid.
3. The stronger the acid or base, the larger the % ionization. 
4. Strong vs weak acid animation 
5. Strong vs weak base animation
3. Section 14.3 H⁺ is really just shorthand for H₃O⁺, the hydronium ion
4. Section 14.4 Water
1. Water can act like both an acid and base depending on its environment = amphiprotic.  Able to lose or gain H⁺.  
2. Water can react with itself (self ionization) as folllows:  H₂O + H₂O  D H₃O⁺ + OH^-      Watch it here. 
3. Pure water doesn't conduct electricity which means there are few ions.  So does the reaction above proceed very far towards the product ions?  No.  This reaction barely goes forward.  Will K be large or small?  K is in fact teeny tiny.  Since we don't include the water (pure liquid) in this K expression and since it is for water we call the equilibrium constant K_w and it = [H⁺][OH^-] = 1.00 x 10^-14.  Remember to put this number in your calculator as 1EE-14.  DO NOT use the 10^{^} button!  Even if we write 10^-14 you must tell your calculator 1EE-14.  K_w is valid for all water solutions at 25^oC !!!  IMPORTANT. 
4. Example:  What is [H⁺]and [OH^-] in pure water?  Well we know K_w = [H⁺][OH^-] = 1.00 x 10^-14. And for every H⁺ (hydronium ion) there is one OH^- (hydroxide ion) so their concentrations must be equal.  So basically x² = 1.00 x 10^-14. Take square roots and get x = 1.00 x 10^-7M which is = [H⁺] which = [OH^-].
5. All aqueous solutions have some hydronium and hydroxide ions. Acidic solutions have more hydronium than hydroxide while basic solutions have more hydroxide than hydronium.  Neutral solutions have equal amounts of each. Figure 14.2
6. Example:  What is the hydronium concentration if the hydroxide concentration is 4.42 x 10^-4M?  Solution:  K_w / [OH^-] = [H⁺].  So 10^-14 / 4.42 x 10^-4M = 2.26 x 10^-11M  This is basic since there is more hydroxide than hydronium. 
7. Try example 14.4 and problems 14.6-7.
5. Section 14.5  pH
1. pH is a logarithmic scale for hydronium ion concentration.  p is just the mathematical function of -log.  So pD = -logD, pA = -logA and so forth.  pH = -log[H⁺], pOH = -log[OH^-], pK = -logK.
2. pH = 7 is neutral, pH < 7 is acidic getting more acidic closer to zero, pH > 7 is basic getting more basic closer to 14. Figure 14.3
3. Since 1.00 x 10^-14.  = [H⁺][OH^-] we can take the negative log of both sides and get:  -log 10^-14 = -log[H⁺] + -log[OH^-] which simplifies to pH + pOH = 14.00
4. As a solution gets more acidic, the pH goes down, the hydronium ion concentration increases, and the hydroxide ion concentration decreases. 
5. Sig Fig:  the number of sig fig in the concentration is the same as the number of decimal places in the pH or pOH. 
6. Given just one of these (pH, pOH, [H⁺] or [OH^-]) you can find the other three!  You must use the four equations bolded in navy above.  First let's make sure you can use your calculator.  Solve these:
1. -log 4.73 x 10^-7 = ???         answer is 6.325
2. 8.88 = -log ?       answer ? = 1.3 x 10^-9
3. See me if you need help and the calculator review on the HELP page. 
7. If pH is 5.24, what is pOH, [OH^-] and [H⁺]?   Answer:  pOH = 14.00 - 5.24 = 8.76.   5.24 = -log[H⁺] so [H⁺] = 5.8 x 10^-6M.   10^-14 / 5.8 x 10^-6 = [OH^-] = 1.7 x 10^-9M.  Check:  8.76 = -log[OH^-] = 1.7 x 10^-9M.
8. If pOH is 2.89, what is pH, [OH^-] and [H⁺]?   Answer:   pH = 11.11, [OH^-] = 1.3 x 10^-3M and [H⁺] = 7.8 x 10^-12M
9. If [OH^-] is 1.11 x 10^-4M, what is pOH, pH and [H⁺]?   Answer:  pOH = 3.955, pH = 10.045 and [H⁺] = 9.02 x 10^-11M
10. If [H⁺] is 3.33 x 10^-3M, what is pOH, pH and [OH^-]?   Answer:   pOH = 11.522, pH = 2.478 and [OH^-] = 3.01 x 10^-12M
11. Note that you can do these in different orders.  Given pH you can find pOH or [H⁺] first.  So you may solve these in different orders than I did which can give way to rounding errors. So your answers may be a little off from mine which is OK as long as they are close.  Given pOH you can find either pH or [OH^-] first.  Given [OH^-] you can find pOH or [H⁺] first.  Given [H⁺] you can find pH or [OH^-] first.  It really doesn't matter.  Remember to punch in 10^-14 in your calculator as "1 EE -14."  Given [OH^-] and [H⁺] you simply take the log and change sign to find pOH and pH respectively.  Given pH or pOH you have to change sign and take the anti-log to find [H⁺] and [OH^-] respectively.
12. Try example 14.5-6 and problems 14.8-9.
6. Section 14.6 pH is measured by indicators, pH paper, litmus paper, and pH meters.
7. Section 14.7 pH of Strong Acids and Bases
1. Calculating pH for a strong acid or base solution - this is easy since all the original acid or base dissociates completely in water. An equilibrium is NOT set up.  
1. Find the pH for 0.089M nitric acid.  Well, all the strong acid will react and make H⁺ and NO₃^-.  So the final [H⁺] is 0.089M.  pH = -log 0.089 = 1.05 (Whew that is quite acidic - don't drink it!!!
2. Find the pH for a 3.33 x 10^-5M potassium hydroxide solution.  Well, all the KOH will dissociate in water making 3.33 x 10^-5M = [OH^-].  pOH = -log 3.33 x 10^-5 = 4.478.  Then 14.000 - 4.478 = pH = 9.522.
2. Remember strong acids and bases in water do not set up equilibrium, they ionize about 100% so the reaction goes forward to completion and is done.
3. Skip the info on metal oxides. Try examples 14.7, 14.8ab, problem 14.10.
8. Section 14.8 Weak acids and K_a
1. It is more difficult to find the pH of a weak acid solution because the reaction is NOT complete and sets up equilibrium with its conjugate base.  In general HA  D  H⁺ + A^-   and K_a = [H⁺][A^-] / [HA], pK_a = -log K_a
2. As the acid gets stronger, [H⁺] increases, pH decreases, % ionization increases and K_a increases
3. Ka value are in Table 14.2
4. It is important to realize that a weak acid is not the same as a dilute acid.  Dilute vs concentrated refers to the molarity.  Strong vs weak refers to the acid's ability to donate a proton. 
9. Section 14.9 Calculations with K_a
1. Given initial [HA] and K_a you should be able to find all [ ] at equilibrium and the pH. (and vice versa)
2. Given initial [HA] and pH, you should be able to find all [ ] at equilibrium and K_a. The pH gives you the [H⁺] at equilibrium and you can fill in the ICE table from that and then solve for K_a.   (and vice versa)
3. Given all [ ] at equilibrium you should be able to find K_a. Just plug them into the K expression. (and vice versa)
4. For any equilibrium problems with a small K, you can try an approximation.  The approximation is this:  when x is added or subtracted to a number, you can try assuming that x will be so small that it will not change that number.  For example:  0.25 - 0.000042 is still just 0.25 when rounded to the correct sig fig.  To test the approximation you take x / [initial] x 100% and if the % is less than 5, we say the approximation is valid.  If the % is greater than 5, the approximation is not valid and you must solve for x using the quadratic formula.
5. Example:  Find the pH of a 0.77M benzoic acid solution given K_a = 6.5 x 10^-5.  First set up an ICE table. 
1. Plub into the K_a expression:  6.5 x 10^-5 = x² / 0.77
2. Solve for x = 7.0746 x 10^-3
3. Check the approximation:  7.0746 x 10^-3 / 0.77 x 100 = 0.9% which is less than 5 so the approximation is good. 
4. Since x is =[H⁺] find the pH = -log 7.0746 x 10^-3 = 2.15 (we need 2 decimal places in the final answer)
6. For more practice problems on weak acids and their equilibrium click here and then click on Weak Acid and Equilibrium on the left side bar. Try examples 14.9-10 and problems 14.12-15.  
10. Section 14.10 % Ionization 
1. % ionization = [H⁺] at equilibrium / [HA] initial x 100
2. What is the percent ionization for the previous example:  7.0746 x 10^-3 / 0.77 x 100 = 0.92%.  Yes, it is the same calculation as the approximation check.
3. For more practice problems on weak acids and their equilibrium click here and then click on Weak Acid and Equilibrium on the left side bar. Try problem 14.16. 
11. Section 14.11 Polyprotic Acids
1. Some acids can lose more than one H⁺.  These are called polyprotic acids.  One of the strong acids can do this - H₂SO₄ can lose 2 H⁺ and we call it diprotic.  Several weak acids can do this:  H₂CO₃ is also diprotic and H₃PO₄ is triprotic. 
2. These acids lose each H⁺ in a step wise manner.  First one H⁺ comes off and that step has its own K_a1 constant.  It gets harder to remove the subsequent H⁺ and that step has its own much smaller K_a2 constant. In general K_a1 > K_a2 > K_a3 etc. because it gets harder and harder to remove additional H⁺ ions from the acid. Table 14.3 has the K_a values for polyprotic acids. 
3. Example:  Find the pH of a 0.25 M sulfurous acid solution.  Given H₂SO₃ has K_a1 = 1.3 x 10^-2 and K_a2 = 6.3 x 10^-8
1. Write the first step reaction:  H₂SO₃ D   HSO₃^- + H⁺
2. Setup the ICE table and plug into K_a1.  The approximation will not work - K_a1 is not that small and the check result is 23%
3. Solve for x using the quadratic formula.  1.3 x 10^-2 = x² / (0.25 - x)
4. We get x = 0.0509 so after the first step [H₂SO₃] = 0.25 - 0.0509 = 0.20M, [HSO₃^-] = [H⁺] = 0.051M
5. Write the second step reaction:  HSO₃^-  D  SO₃^2- + H⁺
6. Setup the ICE table using the values from the first step (ie  [HSO₃^-] and [H⁺] initially = 0.0509M for the second step
7. Solve for x using the approximations since K_a2 is really small.  6.3 x 10^-8 = x(0.0509+x) / (0.0509-x) is the real setup but by approximating we get this 6.3 x 10^-8 = x(0.0509) / (0.0509) = x 
8. Check the approximation and get 0.00012% so it is definitely valid
9. So the equilibrium concentrations are [H₂SO₃] =  0.20M, [HSO₃^-] =  0.0509 - 6.3 x 10^-8 = 0.051M, [H⁺] = 0.0509 + 6.3 x 10^-8 =0.051M, [SO₃^2-] = 6.3 x 10^-8 M (note all final answers need 2 sig dig)  
10. pH = -log 0.051 = 1.29 (need 2 decimal places)
4. Try example 14.11-12 and problems 14.17-18.
5. So the sum reaction of two or more equilibrium reactions has an equilibrium constant K_tot which equals the product of all the K values for each of the  equilibrium reactions.  
6. Example:  equation 1:  H₂CO₃ + H₂O D  H₃O⁺ + HCO₃^- with K₁ = 2.5 x 10^-4.  Now consider equation 2:  HCO₃^- + H₂O  D H₃O⁺ + CO₃^2- with K₂ = 5.6 x 10^-11. What is the overall reaction and the equilibrium constant for the overall reaction?  Answer:  H₂CO₃ + 2H₂O D  2H₃O⁺ + CO₃^2- with K₂ = 1.4 x 10^-14.
12. Section 14.12 Weak Bases and K_b
1. Ammonia, an amine, is a weak base and reacts with water to produce hydroxide ions.  NH₃ + H₂O   D NH₄⁺ + OH^-
2. K_b goes with this reaction and is = [NH₄⁺][OH^-] / [NH₃].  Table 14.4 lists K_b values. 
3. What makes amines basic???  Well the N in amines has a lone pair of electrons (partial negative charge) and that attracts the H from water.  Remember water is polar with the H a little bit positive and the O a little bit negative (likes poles on a magnet - polar bond).  So opposites attract and the amine base attracts the H⁺ from water.  AMAZING 
4. Calculate the pH of a 1.25M methylamine, CH₃NH₂, solution.  K_b = 4.4 x 10^-4. 
1. Write the equation:  CH₃NH₂  +  H₂O  D  CH₃NH₃⁺  +  OH^-
2. Set up your ICE table
3. Plug into K_b = 4.4 x 10^-4 = x² / 1.25 (using the approximation)
4. solve x = 2.3 x 10^-2
5. check the approximation:  2.3 x 10^-2 / 1.25 x 100 = 1.9%  Yeah - it is valid.
6. Now x is [OH^-] in this ICE table, so pOH = -log[OH^-] = 1.64
7. pH = 14 - 1.64 = 12.36 (which makes sense because it is a basic solution)
5. Try example 14.13 and problems 14.19-20  More practice here. 
13. Section 14.13 K_a and K_b
1. For a weak acid K_a = [H⁺][A^-] / [HA]
2. For the conjugate base K_b = [HA][OH^-] / [A^-]
3. Multiply K_aK_b = [H⁺][A^-] [HA][OH^-] / [A^-][HA] = [H⁺][OH^-] which we recognize don't we???  It is K_w.
4. So K_aK_b = K_w  The acid and base MUST BE CONJUGATE PAIRS!  Not just any old acid and any old base. 
5. What is K_a for HCN given that K_b for CN^- is 2.0 x 10^-5?  1.00 x 10^-14 / 2.0 x 10^-5 = K_a = 5.0 x 10^-10
6. Try example 14.14 and problem 14.21.
14. Section 14.14 Salts
1. Salts are inorganic compounds made of ions resulting from acid base reactions.
2. There is a really cool table at the end of this page – look at it.
3. Neutral salts are the result of strong acid strong base reactions are are neutral - they do not react further with water.  Examples:  NaCl, KBr, LiNO₃.  Notice all these can be part of strong acids and strong bases.  
4. Basic Salts - contain a weak acid's conjugate base.  You must recognize the conjugate base which means you need to know the weak acids discussed above.  Examples:  CH₃COONa, LiF, KCN.  These are basic because they contain the conjugate bases CH₃COO^-, F^- and CN^- which come from weak acids.
1. The conjugate base from these salts reacts with water just like any base and gain H⁺ from water creating hydroxide ions.  This process of basic salts reacting with water is called hydrolysis.
2. Percent ionization is called percent hydrolysis for salts and is calculated by [OH^-] at equilibrium / [base] initial x 100%
3. Calculate the pH and % hydrolysis of a 0.555M lithium acetate solution.  First recognize that LiCH₃COO contains a basic ion.  In water the LiCH₃COO g  Li⁺ + CH₃COO^- completely.  So we end up with 0.555M CH₃COO^-.  
4. Now write the base reaction:  CH₃COO^-  + H₂O  D  CH₃COOH  +  OH^- and setup an ICE table.
5. K_b = x² / (0.555 - x)
6. Using K_b = 5.6 x 10^-10  we approximate and get 5.6 x 10^-10 = x² / 0.555
7. Solve for x = 1.76 x 10^-5 = [OH^-]
8. pOH = -log[OH^-] = 4.75
9. pH = 14.00 - 4.75 = 9.25
10. % hydrolysis = 1.76 x 10^-5 / 0.555 x 100 = 0.00317%  
11. Try example 14.16 and problem 14.23
5. Acidic Salts - contain a weak base's conjugate acid.  You must recognize the conjugate acid which means you need to know the weak bases discussed above.  Examples:  NH₄NO₃, NH₄I.  These are acidic because they contain the conjugate acid NH₄⁺ which comes from the weak base ammonia.  Small highly charged cations are also acidic.  They are surrounded by water molecules and pull the water so closely that one of the H's can pop off of a water molecule.  Aluminum is a small cation with a 3+ charge.  In water six water molecules surround it tightly.  One of the six waters can lose an H:  Al(H₂O)₆³⁺  D  Al(H₂O)₅(OH) ²⁺  + H⁺  K_a = 1.4 x 10^-5.  So Al³⁺ ion in water is acidic.  Figure 14.10
1. Percent ionization is called percent hydrolysis for acidic salts also and is calculated by [H⁺] at equilibrium / [acid] initial x 100%
2. The conjugate acid from these salts reacts with water just like any acid and loses H⁺ to water creating H₃O⁺ hydronium ions. 
3. Calculate the pH and % hydrolysis of 1.1 M ammonium iodide solution.  First recognize that NH₄I contains an acidic ion.  In water NH₄I g   NH₄⁺ + I^-  completely.  So we get 1.1M ammonium ion.  
4. Now write the acid reaction   NH₄⁺ + H₂O  D  NH₃ + H₃O⁺  and setup an ICE table. 
5. K_a = x² / (1.1 - x) and continue working the problem just like any other K problem.
6. % hydrolysis will be x / 1.1 x 100%
7. Try example 14.15 and problem 14.22.
6. Skip the part on salts that contain both acidic and basic ions. 
7. Classify the following salts as acidic, basic or neutral. 
1. KBr - neutral
2. NaF - basic
3. LiCN - basic
4. NH₄Cl - acidic
5. Al(NO₃)₃ - acidic
6. NaCH₃COO - basic
7. NaNO₃ - neutral
8. Problem 14.25 a, b, c
15. Section 14.15 Acid Strength Factors (brief coverage)
1. Why is HF a weak acid when HCl, HBr and HI are strong???  Well the small size of F^- allows it to be real close to the H⁺ with very good orbital overlap.  The single bond between H and F is very strong, and thus hard to break.  The single bonds in HCl, HBr, and HI are not near as strong and they all break easily in water.  
2. Carboxylic acids - important in organic chemistry.  They all end with -COOH which looks like
3. The acid has a very polar OH bond with the H a little bit positive and the O a little bit negative.  The H comes off making H⁺ and the conjugate base COO^- which is stabilized by resonance:  the double bond can be shared between both CO bonds.  R is just any hydrocarbon group like methyl.
4. Acetic acid is a carboxylic acid.  Also note that just because a molecule has OH does not make it basic.  Only metal hydroxides are basic. 
16. Section 14.16 Lewis versus Bronsted definitions
1. Bronsted definitions are what we have been doing:  acids donate H⁺ and bases gain H⁺
2. The Lewis definitions are the perspective from the electrons instead of the proton H⁺.  When an acid loses H⁺ the proton is going away but the electron is staying - so an acid is an electron acceptor.  When a base gain a proton H⁺ it must reach out its electron to attract the H⁺ so the base is an electron donor. Ammonia is a good example of a Lewis base - it must donate it's lone pair of electrons to the H⁺ in order to make the new bond forming NH₄⁺. 
3. Examine ammonia in your text as a good example. Try example 14.17 and problems 14.27-28.
17. Just for fun - ACID RAIN - Nonmetal oxides (compound with a nonmetal and oxygen) are acidic.  Examples:  CO₂(g) plus water in the air makes H₂CO₃ carbonic acid.  SO₃(g) plus water in the air makes H₂SO₄ sulfuric acid.  These two acids are responsible for acid rain. 

 

 Table of Acid, Base and Salt Information

 

Chemical	Example reaction	Equili brium?	In the beaker (picture)
Strong acid	HCl(aq) +  H2O(l)  g  Cl- (aq)  +  H3O+(aq)	no	Cl-, H+ (products only)
Weak acid	HF (aq)  +  H2O(l)    D  F- (aq)  +  H3O+ (aq)	yes	HF, F-, H+  (both sides)
Strong base	NaOH(s)   g  Na+ (aq) +  OH-  (aq)       (does not react with water, dissolves in it)	no	Na+, OH- (products only)
Weak base	NH3(aq)  +  H2O(l)    D  OH- (aq)  +  NH4+ (aq)	yes	NH3, OH-, NH4+ (both sides)
Acidic salt	NH4Cl  g  NH4+ (aq) +  Cl-  (aq)       (does not react with water, dissolves in it) then  the non neutral ion (conjugate acid) reacts further with water NH4+ (aq) +  H2O(l)    D  NH3(aq)  +  H3O+ (aq)	yes	NH3, H+ , NH4+ Cl- (both sides)
Basic salt	KF  g  K+ (aq) +  F-  (aq)       (does not react with water, dissolves in it) then  the non neutral ion (conjugate base ) reacts further with water F- (aq)  +  H2O(l)  D    HF (aq)  +  OH-  (aq)	yes	HF, F-, OH-, K+ (both sides)
Neutral salt	NaCl  g  Na+ (aq) +  Cl-  (aq)       (does not react with water, dissolves in it) both ions are neutral so no further reaction	no	Na+, Cl- (products only)