Chapter 15 Review Acids and Bases Section 15-1

Chapter 15 Review:  Acids and Bases

1. Section fifteen.1 Bronsted Acids and Bases
1. Acids - molecules that can lose H⁺ (proton donors) making H₃O⁺ in h2o (acids lose H⁺)
2. Bases - molecules than can proceeds H⁺ (proton acceptors) and/or brand OH^- in water (bases proceeds H⁺ / make OH^-)
3. Cohabit acrid/base pairs - differ by one H⁺
1. when an acrid loses ane H⁺ it becomes its conjugate base of operations
2. when a base gains 1 H⁺ it becomes its conjugate acid
3. examples (acrid get-go then conjugate base of operations) H₃PO₄ and H_iiPO_four ^-,  NH₄ ⁺ and NH₃, H₃O⁺ and H₂O, H₂PO_iv ^- and HPO₄ ^2-, H₂O and OH^-, CH₃COOH and CH₃COO^-
4. Discover that after losing H⁺ the charge goes down past 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 acrid (ca) ad conjugate base (cb) Answer:  H₂CO_three(aq) is the acrid (carbonic acid),  H₂O(50) acts as the base here,   HCO₃ ^-(aq) is the cb,  H₃O⁺(aq) is the ca
2. CH₃NH₂ (aq) + H_iiO(l) D   CH₃NH₃ ⁺ (aq) + OH^-(aq).  Label them as acrid, base, conjugate acrid (ca) advertizement conjugate base (cb) Answer:  CH_iiiNH_two (aq) is the base of operations (an amine),  H₂O(l) acts as the acid here,   CH₃NH₃ ⁺ (aq) is the ca, OH^-(aq) is the cb
3. Retrieve that H⁺ is but brusque hand for H₃O⁺called hydronium ion.  H⁺ doesn't really exist in h2o.
4. Remember that water is non an acrid or base, information technology tin just act like ane depending on the other molecules around it.
5. Try instance 15.i - 2, bug 15.1 - 15.three
2. Section 15.ii Acid and Base of operations Forcefulness
1. Predicting a direction to attain equilibrium.  You lot need to refer to a Table that has acids and conjugate bases in order of forcefulness like Table 15.1 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 than the weaker acid so the equilibrium will lie on the weaker acid side.
1. Consider this reaction:  CH₃COOH + So₄ ^2- D   CH₃COO^- + HSO_iv ^-.  There are two acids here:  acetic acid CH₃COOH and HSO₄ ^- ion.  Which is stronger???  HSO_four ^- is stronger and then it will react more, and it reacts towards the left side (reactants).  And then the equilibrium will lie on the reactant side - few products will be made.  We will by and large have acetic acid and sulfate ion in the beaker with only a bit of acetate ion and hydrogen sulfate ions in the beaker. K is less than one, G is a small number, reactants are favored.
2. Consider this reaction:  NH₃ + HF D   NH_iv ⁺ + F^-.  The two acids are HF and NH_iv ⁺ ion.  Which is stronger?  HF is stronger so it volition react more and it reacts towards the products.  The equilibrium lies on the production side. K is greater than one, products are favored.
3. Endeavour example 15.3 and problems 15.4-5.
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 manner - no equilibrium is set up.
1. Memorize the vi strong acids:  muriatic acid HCl, hydrobromic acrid HBr, hydroiodic acrid Hello, nitric acid HNO₃, perchloric acid HClO_four and sulfuric acrid, H_twoAnd so_four.
2. All of them follow this pattern in h2o:  HCl (aq) + H_twoO (l) m   Cl^- (aq) + H_threeO⁺ (aq) which tin can besides be written as just HCl (aq) yard   Cl^- (aq) + H⁺ (aq)
3. If I begin with 2.5M strong acid, I will have ii.5M hydronium ion and 2.5M conjugate base.
4. A flick of HCl in water would have no HCl molecules, it would have only the ions since it all ionized.
3. Weak acids lose but a small per centum of H⁺ and only partially ionize in water making hydronium ions. They set up equilibrium.
1. All other acids are weak except the vi potent ones.  Four you should be familiar with are acetic acid CH_threeCOOH, phosphoric acid H₃PO_iv, carbonic acid H₂CO_iii and hydrofluoric acrid HF.
2. All weak acids follow this pattern in water:  HF (aq) + H_iiO (fifty) D   F^- (aq) + H₃O⁺ (aq) which can also be written every bit just HF (aq) D   F^- (aq) + H⁺ (aq)
3. If I brainstorm with 2.5M weak acid, I do non know how much hydronium ion and cohabit base I accept without computing and Ice tabular array.
4. A pic of HF in water would have mostly HF molecules with few ions.
4. Stiff bases well-nigh 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 deliquesce and dissociate virtually completely in water merely like:  NaOH (s) g Na⁺(aq) + OH^-(aq)
3. If I brainstorm 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 deliquesce and dissociate to a small extent.  Ammonia and other amines are weak bases because they react WITH water making hydroxide.
2. Other metallic hydroxides similar Mg(OH)_two (s) D Mg⁺²(aq) + ii OH^-(aq)
3. Amines react with water like:  NH₃ + H_twoO D   NH_four ⁺ + OH^-
4. If I begin with ane.0M ammonia, I don't know how much hydroxide I have without computing.
5. A picture of ammonia in h2o has more often than not NH₃ molecules with some ammonium and hydroxide ions.
6. Notes
1. A potent acid'southward cohabit base of operations is really not basic simply 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 cohabit acid.
3. The stronger the acid or base of operations, the larger the % ionization.
4. Strong vs weak acrid animation
5. Stiff vs weak base of operations animation
3. Section 15.3 H⁺ is really just autograph for H₃O⁺, the hydronium ion
4. Department 15.4 Water
1. H2o tin can act like both an acid and base depending on its surroundings = amphiprotic.  Able to lose or proceeds H⁺.
2. H2o tin react with itself (cocky ionization) as folllows:  H₂O + H₂O D H₃O⁺ + OH^- Watch it hither.
3. Pure water doesn't conduct electricity which means there are few ions.  So does the reaction higher up proceed very far towards the product ions?  No.  This reaction barely goes forrad.  Will K be big or minor?  K is in fact teeny tiny.  Since we don't include the h2o (pure liquid) in this Yard expression and since it is for water we call the equilibrium constant G_w and it = [H⁺][OH^-] = 1.00 x 10^-14 . Remember to put this number in your reckoner every bit 1EE-fourteen.  Practise Not use the 10^{^} button! Fifty-fifty if nosotros write ten^-fourteen you must tell your calculator 1EE-fourteen.  K_w is valid for all water solutions at 25^oC !!!  IMPORTANT.
4. Instance:  What is [H⁺]and [OH^-] in pure water?  Well we know M_{due west} = [H⁺][OH^-] = 1.00 x 10^-xiv. And for every H⁺ (hydronium ion) there is one OH^- (hydroxide ion) then their concentrations must be equal.  So basically x² = i.00 x 10^-xiv. Have square roots and get 10 = one.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 bones solutions have more hydroxide than hydronium.  Neutral solutions take equal amounts of each.Effigy 15.2
6. Example:  What is the hydronium concentration if the hydroxide concentration is 4.42 x x^-41000?  Solution:  K_w / [OH^-] = [H⁺].  And so 10^-14 / iv.42 x 10^-fourM = ii.26 x x^-11M  This is basic since there is more hydroxide than hydronium.
7. Attempt instance fifteen.4 and problems xv.half dozen-7.
5. Department 15.5  pH
1. pH is a logarithmic calibration for hydronium ion concentration.  p is just the mathematical part of -log.  So pD = -logD, pA = -logA then forth. pH = -log[H⁺] , pOH = -log[OH^-], pK = -logK.
2. pH = seven is neutral, pH < seven is acidic getting more acidic closer to naught, pH > seven is bones getting more basic closer to 14.Figure xv.three
3. Since ane.00 x 10^-14.  = [H⁺][OH^-] we can take the negative log of both sides and go:  -log ten^-14 = -log[H⁺] + -log[OH^-] which simplifies to pH + pOH = 14.00
4. Equally 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 only one of these (pH, pOH, [H⁺] or [OH^-]) you tin can observe the other iii! You must use the four equations bolded in navy to a higher place.  First let'southward make sure you tin use your calculator.  Solve these:
1. -log 4.73 x 10^-vii = ???         reply is half-dozen.325
2. 8.88 = -log ?       answer ? = 1.3 10 x^-9
3. See me if you need help and the calculator review on the HELP folio.
7. If pH is 5.24, what is pOH, [OH^-] and [H⁺]?   Answer:  pOH = fourteen.00 - 5.24 = eight.76.   5.24 = -log[H⁺] so [H⁺] = 5.8 x x^-6M.   ten^-14 / v.8 x 10^-six = [OH^-] = i.seven ten 10^-ixK.  Check:  8.76 = -log[OH^-] = i.seven x ten^-9Thousand.
8. If pOH is ii.89, what is pH, [OH^-] and [H⁺]?   Answer:   pH = 11.11, [OH^-] = 1.3 x 10^-3M and [H⁺] = 7.eight x x^-12Thousand
9. If [OH^-] is 1.eleven x 10^-4M, what is pOH, pH and [H⁺]?   Respond:  pOH = three.955, pH = 10.045 and [H⁺] = ix.02 10 10^-xiM
10. If [H⁺] is three.33 x ten^-31000, what is pOH, pH and [OH^-]?   Reply:   pOH = 11.522, pH = 2.478 and [OH^-] = three.01 ten 10^-12M
11. Note that yous can practice these in different orders.  Given pH you lot can find pOH or [H⁺] first.  So you may solve these in different orders than I did which tin requite fashion to rounding errors. So your answers may be a little off from mine which is OK every bit long equally they are close.  Given pOH you can find either pH or [OH^-] outset.  Given [OH^-] you can find pOH or [H⁺] first.  Given [H⁺] you tin can observe pH or [OH^-] kickoff.  It actually doesn't matter.  Remember to dial in 10^-14 in your reckoner equally "1 EE -14."  Given [OH^-] and [H⁺] you simply take the log and change sign to discover pOH and pH respectively. Given pH or pOH you have to change sign and accept the anti-log to find [H⁺] and [OH^-] respectively.
12. Effort example xv.5-half dozen and problems fifteen.eight-9.
6. Section 15.half dozen pH is measured by indicators, pH newspaper, litmus paper, and pH meters.
7. Department 15.7 pH of Strong Acids and Bases
1. Computing pH for a stiff acid or base solution - this is easy since all the original acid or base dissociates completely in h2o. An equilibrium is Not ready.
1. Discover the pH for 0.089M nitric acid.  Well, all the potent acid volition react and brand H⁺ and NO₃ ^-.  Then the last [H⁺] is 0.089M.  pH = -log 0.089 = one.05 (Whew that is quite acidic - don't drinkable it!!!
2. Observe 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 xiv.000 - 4.478 = pH = 9.522.
2. Remember stiff acids and bases in water exercise not set up equilibrium, they ionize about 100% so the reaction goes forwards to completion and is done.
3. Skip the info on metal oxides. Endeavor examples 15.vii, xv.8ab, trouble 15.10.
8. Section 15.8 Weak acids and 1000_a
1. It is more than difficult to find the pH of a weak acrid solution because the reaction is NOT complete and sets up equilibrium with its cohabit base.  In general HA D H⁺ + A^-   and 1000_a = [H⁺][A^-] / [HA], pK_a = -log K_a
2. As the acid gets stronger, [H⁺] increases, pH decreases, % ionization increases and Thou_a increases
3. Ka value are in Table 15.2
4. Information technology is of import to realize that a weak acid is non the aforementioned as a dilute acid.  Dilute vs full-bodied refers to the molarity.  Strong vs weak refers to the acrid's power to donate a proton.
9. Section fifteen.9 Bug with K_a
1. Given initial [HA] and K_a you should be able to find all [ ] at equilibrium and the pH.
2. Given initial [HA] and pH, you should be able to notice all [ ] at equilibrium and Grand_a. The pH gives you the [H⁺] at equilibrium and y'all can fill in the ICE table from that and then solve for Chiliad_a.
3. Given all [ ] at equilibrium y'all should be able to discover G_a. Just plug them into the K expression.
4. For whatsoever equilibrium problems with a small K, you can endeavour 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 non modify that number.  For example:  0.25 - 0.000042 is notwithstanding simply 0.25 when rounded to the correct sig fig.  To test the approximation you accept 10 / [initial] ten 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:  Observe the pH of a 0.77M benzoic acid solution given Chiliad_a = 6.5 x 10^-5.  First gear up upwards an Ice table.
1. Plub into the K_a expression:  six.5 x 10^-five = ten^two / 0.77
2. Solve for ten = seven.0746 10 10^-3
3. Cheque the approximation:  7.0746 x x^-3 / 0.77 x 100 = 0.9% which is less than v so the approximation is good.
4. Since x is =[H⁺] find the pH = -log 7.0746 x 10^-iii = 2.15 (nosotros need 2 decimal places in the final answer)
6. For more practice issues on weak acids and their equilibrium click hither and and so click on Weak Acid and Equilibrium on the left side bar. Attempt examples fifteen.nine-10 and problems xv.12-15.
10. Section xv.10 % Ionization
1. % ionization = [H⁺] at equilibrium / [HA] initial x 100
2. What is the percent ionization for the previous example:  seven.0746 10 ten^-3 / 0.77 ten 100 = 0.92%. Yes, it is the same calculation equally the approximation bank check.
3. For more practice issues on weak acids and their equilibrium click here and and so click on Weak Acid and Equilibrium on the left side bar. Try problem 15.xvi.
11. Section xv.xi Polyprotic Acids
1. Some acids can lose more than one H⁺.  These are called polyprotic acids.  One of the strong acids can practice this - H_iiAnd then₄ can lose 2 H⁺ and we call it diprotic.  Several weak acids can practice this:  H_twoCO_three is also diprotic and H₃PO₄ is triprotic.
2. These acids lose each H⁺ in a footstep wise manner.  Get-go one H⁺ comes off and that pace has its ain M_a1 constant.  Information technology gets harder to remove the subsequent H⁺ and that footstep has its ain much smaller Thousand_a2 constant. In general K_a1 > K_a2 > Chiliad_a3 etc. considering it gets harder and harder to remove boosted H⁺ ions from the acrid. Table 15.3 has the K_a values for polyprotic acids.
3. Example:  Find the pH of a 0.25 M sulfurous acrid solution.  Given H₂So₃ has Yard_a1 = 1.3 x 10^-2 and K_a2 = six.3 x 10^-viii
1. Write the showtime stride reaction:  H₂SO₃ D    HSO₃ ^- + H⁺
2. Setup the ICE tabular array and plug into K_a1.  The approximation volition not work - K_a1 is not that small and the check issue is 23%
3. Solve for x using the quadratic formula.  1.3 x 10^-2 = x² / (0.25 - x)
4. We become x = 0.0509 so after the commencement step [H₂And so_three] = 0.25 - 0.0509 = 0.20M, [HSO₃ ^-] = [H⁺] = 0.051M
5. Write the second stride reaction:  HSO₃ ^- D   SO₃ ^two- + H⁺
6. Setup the ICE tabular array using the values from the kickoff step (ie  [HSO₃ ^-] and [H⁺] initially = 0.0509M for the second step
7. Solve for 10 using the approximations since K_a2 is actually modest.  6.three x 10^-8 = x(0.0509+10) / (0.0509-ten) is the real setup but past approximating we become this 6.3 x ten^-8 = x(0.0509) / (0.0509) = ten
8. Check the approximation and go 0.00012% so it is definitely valid
9. So the equilibrium concentrations are [H₂And then_three] =  0.20M, [HSO₃ ^-] =  0.0509 - 6.3 x 10^-8 = 0.051M, [H⁺] = 0.0509 + half-dozen.3 x 10^-viii =0.051M, [Then₃ ^2-] = vi.3 ten x^-8 M (note all final answers need 2 sig dig)
10. pH = -log 0.051 = one.29 (need 2 decimal places)
4. Attempt case fifteen.eleven and problems 15.17-18.
5. So the sum reaction of two or more equilibrium reactions has an equilibrium abiding K_tot which equals the product of all the 1000 values for each of the  equilibrium reactions.
6. Example:  equation 1:  H₂CO₃ + H_iiO D   H_iiiO⁺ + HCO_iii ^- with K_i = 2.5 x 10^-4.  Now consider equation two:  HCO₃ ^- + H₂O D H₃O⁺ + CO_three ^ii- with Thousand₂ = 5.6 x x^-11. What is the overall reaction and the equilibrium abiding for the overall reaction?  Answer:  H₂CO_iii + 2H₂O D   2H₃O⁺ + CO₃ ^2- with K_ii = one.four x 10^-14.
12. Section 15.12 Weak Bases and Thou_b
1. Ammonia, an amine, is a weak base and reacts with water to produce hydroxide ions.  NH₃ + H_twoO D NH₄ ⁺ + OH^-
2. K_b goes with this reaction and is = [NH_four ⁺][OH^-] / [NH₃]. Table xv.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 h2o.  Call back water is polar with the H a little bit positive and the O a little bit negative (likes poles on a magnet - polar bail).  So opposites attract and the amine base of operations attracts the H⁺ from h2o.  AMAZING
4. Calculate the pH of a i.25M methylamine, CH₃NH_two, solution.  G_b = 4.4 x 10^-4.
1. Write the equation:  CH₃NH_two  +  H₂O D   CH_iiiNH₃ ⁺  +  OH^-
2. Set up your ICE table
3. Plug into K_b = four.4 x 10^-4 = 10^two / 1.25 (using the approximation)
4. solve 10 = two.3 x x^-2
5. check the approximation:  ii.3 x 10^-two / 1.25 ten 100 = 1.ix%  Yeah - information technology is valid.
6. Now x is [OH^-] in this Ice table, and then pOH = -log[OH^-] = i.64
7. pH = xiv - i.64 = 12.36 (which makes sense because information technology is a bones solution)
5. Try case 15.12 and problems fifteen.19-twenty More practice here.
13. Section 15.xiii 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 Yard_aK_b = [H⁺][A^-] [HA][OH^-] / [A^-][HA] = [H⁺][OH^-] which we recognize don't nosotros???  It is 1000_w.
4. So Thou_aK_b = K_westward  The acrid and base MUST Be CONJUGATE PAIRS!  Not merely any quondam acid and whatever old base.
5. What is K_a for HCN given that Thousand_b for CN^- is ii.0 x ten^-5?  1.00 10 10^-fourteen / 2.0 x ten^-5 = One thousand_a = five.0 10 10^-10
6. Try example 15.13 and problem 15.21.
14. Section xv.14 Salts
1. Salts are inorganic compounds fabricated of ions resulting from acid base reactions.
2. There is a really absurd tabular array at the cease of this page � look at information technology.
3. Neutral salts are the consequence of potent acid strong base of operations reactions are are neutral - they exercise not react further with water.  Examples:  NaCl, KBr, LiNO₃.  Discover all these can exist part of strong acids and stiff bases.
4. Basic Salts - comprise a weak acid'due south conjugate base.  You must recognize the conjugate base which means you lot need to know the weak acids discussed above.  Examples:  CH₃COONa, LiF, KCN.  These are basic because they contain the cohabit bases CH₃COO^-, F^- and CN^- which come up from weak acids.
1. The conjugate base from these salts reacts with water just similar any base and gain H⁺ from water creating hydroxide ions.  This process of basic salts reacting with water is chosen hydrolysis.
2. Percent ionization is chosen percent hydrolysis for salts and is calculated by [OH^-] at equilibrium / [base] initial 10 100%
3. Summate the pH and % hydrolysis of a 0.555M lithium acetate solution.  Outset recognize that LiCH₃COO contains a bones ion.  In water the LiCH₃COO thousand   Li⁺ + CH₃COO^- completely.  So we stop up with 0.555M CH₃COO^-.
4. Now write the base reaction:  CH₃COO^-  + H₂O D   CH₃COOH  +  OH^- and setup an ICE tabular array.
5. K_b = 10ⁱⁱ / (0.555 - x)
6. Using Grand_b = 5.6 10 10^-10 we approximate and get 5.6 x 10^-10 = x² / 0.555
7. Solve for x = 1.76 ten 10^-5 = [OH^-]
8. pOH = -log[OH^-] = 4.75
9. pH = fourteen.00 - 4.75 = 9.25
10. % hydrolysis = 1.76 x 10^-5 / 0.555 x 100 = 0.00317%
11. Try example 15.15 and problem fifteen.23
5. Acidic Salts - contain a weak base of operations's conjugate acid.  Y'all must recognize the conjugate acid which means yous need to know the weak bases discussed in a higher place.  Examples:  NH_ivNO₃, NH_fourI.  These are acidic considering they contain the cohabit acid NH_iv ⁺ which comes from the weak base of operations ammonia.  Small-scale highly charged cations are besides acidic.  They are surrounded past water molecules and pull the h2o and then closely that ane of the H's tin pop off of a water molecule.  Aluminum is a small cation with a 3+ charge.  In water six water molecules surround information technology tightly.  1 of the half dozen waters tin lose an H:  Al(H₂O)₆ ³⁺ D   Al(H₂O)_v(OH) ²⁺  + H⁺  K_a = one.four x ten^-v.  And then Al³⁺ ion in water is acidic.
1. Percent ionization is called percentage hydrolysis for acidic salts also and is calculated by [H⁺] at equilibrium / [acid] initial ten 100%
2. The conjugate acid from these salts reacts with water simply like any acid and loses H⁺ to h2o 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 h2o NH₄I chiliad    NH_four ⁺ + I^-  completely.  And so we get 1.1M ammonium ion.
4. At present write the acid reaction   NH₄ ⁺ + H_twoO D   NH₃ + H_threeO⁺  and setup an Ice tabular array.
5. Chiliad_a = x² / (1.1 - ten) and go along working the problem only similar whatever other Thou problem.
6. % hydrolysis volition exist x / 1.1 ten 100%
7. Try instance 15.14 and problem 15.22.
6. Classify the post-obit salts every bit acidic, basic or neutral.
1. KBr - neutral
2. NaF - basic
3. LiCN - basic
4. NH₄Cl - acidic
5. Al(NO_iii)₃ - acidic
6. NaCH_iiiCOO - basic
7. NaNO₃ - neutral
15. Section 15.15 Acrid Force Factors (brief coverage)
1. Why is HF a weak acid when HCl, HBr and Howdy are stiff???  Well the small size of F^- allows information technology to be real close to the H⁺ with very practiced orbital overlap.  The single bond between H and F is very strong, and thus difficult to break.  The unmarried bonds in HCl, HBr, and HI are non virtually as stiff and they all break easily in water.
2. Carboxylic acids - of import in organic chemistry.  They all end with -COOH which looks like
3. The acid has a very polar OH bond with the H a picayune fleck positive and the O a trivial fleck negative.  The H comes off making H⁺ and the cohabit base COO^- which is stabilized by resonance:  the double bond can be shared between both CO bonds.  R is but any hydrocarbon group like methyl.
4. Acetic acid is a carboxylic acid.  Too note that just because a molecule has OH does not arrive basic.  Only metal hydroxides are basic.
16. Section 15.12 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 acrid loses H⁺ the proton is going away but the electron is staying - and so an acid is an electron acceptor.  When a base gain a proton H⁺ it must reach out its electron to concenter the H⁺ so the base is an electron donor. Ammonia is a proficient example of a Lewis base - it must donate it's lone pair of electrons to the H⁺ in lodge to make the new bail forming NH_four ⁺.
3. Examine ammonia in your text as a good instance. Endeavor example fifteen.16 and problems xv.27-28.
17. Just for fun - ACID Pelting - Nonmetal oxides (compound with a nonmetal and oxygen) are acidic.  Examples:  CO_ii(g) plus water in the air makes H₂CO₃ carbonic acid.  SO_three(g) plus water in the air makes H₂So_four sulfuric acid.  These two acids are responsible for acid rain.

Table of Acid, Base and Salt Information

Chemical	Example reaction	Equili brium?	In the chalice (picture)
Strong acid	HCl(aq) +� HiiO(l) � one thousand � Cl- (aq)� +� HiiiO+ (aq)	no	Cl-, H+ (products but)
Weak acrid	HF (aq) � +� H2O(l) ��� D � F- (aq)� +� HthreeO+ (aq)	yes	HF, F-, H+ � (both sides)
Stiff 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(fifty) ��� D � OH- (aq)� +� NH4 + (aq)	yes	NHiii, OH-, NH4 + (both sides)
Acidic salt	NHfourCl� g � NHfour + (aq) +� Cl- � (aq)������ (does non react with h2o, dissolves in it) and so� the non neutral ion (conjugate acrid) reacts further with h2o NH4 + (aq) +� H2O(50) ��� D � NHiii(aq) � +� HiiiO+ (aq)	yes	NH3, H+ , NH4 + Cl- (both sides)
Basic salt	KF� g � K+ (aq) +� F- � (aq)������ (does not react with water, dissolves in it) and then� the non neutral ion (conjugate base ) reacts further with water F- (aq)� +� H2O(l) � D ��� HF (aq) � +� OH- � (aq)������	yes	HF, F-, OH-, G+ (both sides)
Neutral salt	NaCl� 1000 � Na+ (aq) +� Cl- � (aq)������ (does not react with h2o, dissolves in it) both ions are neutral so no further reaction	no	Na+, Cl- (products only)

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Source: https://web.gccaz.edu/~kimld88531/152Reviews/152rev15.htm

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