 acids always have as the state and always have A. Naming Acids and Bases 6.1 Theories of Acids and Bases Chapter 6: Acids & Bases Rules 1. hydrogen.

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 acids always have as the state and always have A. Naming Acids and Bases 6.1 Theories of Acids and Bases Chapter 6: Acids & Bases Rules 1. hydrogen becomes acid 2. hydrogen becomes acid 3. hydrogen becomes acid ____ide hydr____ic _____ate_____ic ____ite____ous (aq) hydrogen **Hydro and his side kick were skating, when they ate shit on ice, and became a little less rite chous.

Examples: hydroiodic acid phosphoric acid nitrous acid sulphurous acid Change each of the following to the appropriate acid name and give the formula: HI(aq) H 3 PO 4 (aq) HNO 2 (aq) H 2 SO 3 (aq) 1. hydrogen iodide = 2. hydrogen phosphate = 3. hydrogen nitrite = 4. hydrogen sulphite =

 most bases are ionic compounds that are named accordingly Examples: sodium hydroxide sodium hydrogen carbonate magnesium hydroxide ammonia Name each of the following bases: 1. NaOH(aq) = 2. NaHCO 3 (aq) = 3. Mg(OH) 2 (aq) = 4. NH 3 (aq) =

 IUPAC names for acids and bases are simply the word “aqueous” followed by the ionic name Examples: aqueous hydrogen iodide aqueous magnesium hydroxide aqueous hydrogen sulphite aqueous sodium hydrogen carbonate Write the IUPAC name for each of the following acids and bases: 1. hydroiodic acid = 2. magnesium hydroxide = 3. sulphurous acid = 4. sodium hydrogen carbonate =

B. Properties of Acids and Bases  are of a substance empirical propertiesobservable properties  acids, bases and neutral substances have some properties that distinguish them and some that are the same

AcidsBases Neutral Substances               sour bitter electrolytes electrolytes, non- electrolytes bases acids indicators do not H 2(g) eg)HCl(aq), H 2 SO 4 (aq) eg) NaCl(aq), Pb(NO 3 ) 2 (aq) eg)Ba(OH) 2 (aq) NH 3 (aq) less than 7 greater than 7 of 7 litmus - bromothymol blue - redblue yellow taste neutralize react with react with to produce metals phenolphthalein - colourless pink affect indicators the same way pH

 first proposed theory on acids and bases  his theory was that some compounds form  his explanation of the properties of acids and bases is called the electrically charged particles Svante Arrhenius Arrhenius theory of acids and bases C. Arrhenius Definition when in solution

 an Arrhenius is a substance that to form in water  a will in an aqueous solution baseincrease the [OH - (aq)] basedissociates hydroxide ions, OH  (aq),  an Arrhenius is a substance that (because it is molecular) to form  an will in an aqueous solution acidincrease the [H + (aq)] acidionizes hydrogen ions, H + (aq), in water

D. Modified Arrhenius Definition  the original definition of acids and bases proposed by Arrhenius is good but it has  some substances that might be predicted to be are actually limitations neutral basic  it has been found that not all bases contain the eg)Na 2 CO 3 (aq), NH 3 (aq) hydroxide ionas part of theirchemical formula

 an Arrhenius is a substance that in aqueous solution base (modified) reacts with water NH 3 (aq) eg) OH  (aq)+NH 4 + (aq)  H 2 O( )+ to produce OH  (aq) ions

 it has been found using analytical technology like X-ray crystallography that in an aqueous solution  when acids ionize, they produceH + (aq) eg) HCl(g)  H + (aq) + Cl  (aq) H + (aq) ions do not exist in isolation  the hydrogen ion is extremely positive in charge and water molecules themselves are very polar so… it is that would exist in water without being attracted to the of other highly unlikelyhydrogen ions negative poleswater molecules

H 3 O + (aq)  this results in the formation of thehydronium ion + Hydrogen ion Water molecule

 an Arrhenius is a substance that in aqueous solution acid (modified) reacts with water HCl(aq) eg) H 3 O + (aq)+Cl  (aq)  H 2 O( )+ H 2 SO 3 (aq)H 3 O + (aq)+HSO 3  (aq)  H 2 O( )+ to produceH 3 O + (aq) ions

Review: Definitions of Acids and Bases Initial Arrhenius Acid and Base  Acids dissociate to form H+ ions in a solution.  Ex:  HCl (aq) = H + (aq) + Cl - (aq)  Bases dissociate to form OH- ions in a solution.  NaOH (aq) = OH - (aq) + Na + (aq)  Wrong for bases that don’t finish with OH – doesn’t explain a few bases…. Modified Arrhenius Acid and Base  Acids react with water to form H 3 O + ions in a solution.  Ex:  Bases react with water to form OH- ions in a solution. (Only for bases that don’t finish with OH)

Practice a few!!!  Dissociate these acids and bases  HNO 3 (aq) + H 2 O (l)  HCOOH (aq) + H 2 O (l)

 the of a substance depend on two things: acidic and basic properties 1.the of the solution 2.the of the acid or base 6.2 Strong and Weak Acids and Bases concentration identity

A. Strong Acids and Weak Acids  an acid that ionizes almost in water is called a  100% of the becomes strong acid 100% eg) HCl(aq) + H 2 O( )  Cl  (aq)+H 3 O + (aq) HCl(aq) H 3 O + (aq) and Cl  (aq)  the concentration of the is the as the concentration of the it came from H 3 O + (aq) same acid  strong acids are strong electrolytes and react vigorously with metals

 there are 6 strong acids: HNO 3 (aq) H 2 SO 4 (aq) HCl(aq) HI(aq) HBr(aq) HClO 4 (aq) nitric acid sulfuric acid hydrochloric acid hydroiodic acid hydrobromic acid perchloric acid ***on your periodic table

 a and only a small percentage of the acid forms weak acid does not ionize 100% react much less vigorously with metals ions in solution  we use the for weak acids eg) CH 3 COOH(aq) + H 2 O( ) ⇌ CH 3 COO  (aq) +H 3 O + (aq) equilibrium arrow  weak acids are weak electrolytes and

B. Strong Bases and Weak Bases  a base that dissociates into ions in water is called a  are strong bases strong base 100%  a and only a small percentage of the base forms  we use the for weak bases ionic hydroxides and metallic oxides eg) NaOH(aq)  OH  (aq)+Na + (aq) ions in solution weak base does not dissociate 100% eg) NH 3 (aq) + H 2 O( ) ⇌ OH  (aq) + NH 4 + (aq) equilibrium arrow

C. Monoprotic and Polyprotic Acids  acids that have only per molecule that can are called  monoprotic acids can be eg)CH 3 COOH(aq)HNO 3 (aq),HF(aq),HCl(aq), strong or weak one hydrogen atom ionize monoprotic acids

 acids that contain that can are called eg)H 3 PO 4 (aq)H 2 SO 4 (aq),  acids with are, with are two hydrogensdiprotic three hydrogenstriprotic two or more hydrogen atoms ionizepolyprotic acids

 when polyprotic acids ionize, only hydrogen is removed at a time, with each acid becoming one progressively weaker eg) HSO 4  (aq) + H 3 O + (aq) SO 4 2  (aq)+H 3 O + (aq)  H 2 O( ) +  + H 2 SO 4 (aq)

D. Monoprotic and Polyprotic Bases  bases that are called  bases that react with water in are called eg) NaOH(s) react with water in only one step to form hydroxide ions monoprotic bases two or more steps polyprotic bases eg) CO 3 2  (aq), PO 4 3  (aq) ***complex ions with more than 1- charge!!!

 as with polyprotic acids, only eg) HCO 3  (aq) + OH  (aq) H 2 CO 3 (aq)+OH  (aq)  H 2 O( ) +  + CO 3 2  (aq) one OH  (aq) is formed at a time, and each new base formed is than the last weaker

E. Neutralization  the reaction between an acid and a base produces an  the products of are both ionic compound and water neutralneutralization  in a neutralization reaction or between a, the product is always water strong acid and a strong base acid-base reaction H 3 O + (aq)2 H 2 O( )  OH  (aq)+ wateracid+base → a salt+ eg) HCl(aq) + KOH(aq) → KCl(aq) + HOH( )

F. Acid and Base Spills  there are many uses for both acids and bases in our households and in industry  due to their, special care must be used when they are being reactivity and corrosiveness produced and transported

 the two ways to deal with acid or base spills are: 1. dilution: 2. neutralization: reduce the by adding you always use a for the neutralization so you aren’t left with another hazardous situation concentration water weak acid or base

A. Ion Concentration in Water  the “self-ionization” of water is verysmall (only 2 in 1 billion)  H 3 O + (aq)OH - (aq) + + H 2 O( ) hydronium ions hydroxide ions equal and constant in pure water [H 3 O + (aq)]= [OH - (aq)] = 1.0 x mol/L  the concentration of and are 6.3 Acids, Bases and pH

B. The pH Scale  in 1909, Soren Sorenson devised the  it is used because the [H 3 O + (aq)] is pH scale very small  at 25  C (standard conditions), most solutions have a pH that falls between 0.0 and  it is possible to have a pH and a pH  it is a based on whole numbers that are powers of 10 logarithmic scale 14.0 negative above 14

 there is a for every change in on the pH scale 10-fold change in [H 3 O + (aq)] a solution with a pH of 11 is times more basic than a solution with a pH of 9 10  10 = 100 pH Scale 0714 more acidicmore basic neutral 1 eg)

Try These: 1. [H 3 O + (aq)] = 1 x mol/LpH = 2. [H 3 O + (aq)] = 1.0 x mol/LpH = 3. [H 3 O + (aq)] = 6.88 x mol/LpH = 4. [H 3 O + (aq)] = 9.6 x mol/LpH = pH =  log [H 3 O + (aq)] C. Calculating pH and pOH ***New sig dig rule: when reporting pH or pOH values, only the numbers to the count as significant right of the decimal place

Example 6.30 g of HNO 3 is dissolved in 750 mL of water. What is the pH ? HNO 3 (aq) m = 6.30 g M = g/mol V = L c = 0.133…mol/L x 1/1 = 0.133…mol/L pH = -log[H + (aq)] = -log[0.133… mol/L] = NO 3 - (aq)H + (aq)  n = m M = 6.30 g g/mol = …mol c = n V = …mol L = 0.133…mol/L

 just as deals with deals with ***p just means  log pH [H 3 O + (aq)], pOH [OH  (aq)]  at SATP… pH + pOH = 14 pOH pH

 to calculate the use the same formulas as pH but substitute the pOH, [OH  (aq)] pOH =  log[OH  (aq)] Try These: 1. [OH  (aq)] = 1.0  mol/LpOH = 2. [OH  (aq)] = 6.22  mol/LpOH = 3. [OH  (aq)] =  mol/LpOH = 4. [OH  (aq)] = 2  mol/LpOH =

Summarizing pH and POH  pH – concentration of H+ in your solution. 0 = strong acid, 7 = neutral, 14 = strong base  pOH – concentration of OH- in your solution. 14 = strong acid, 7 neutral, 0 = strong base pH =  log [H 3 O + (aq)] pOH =  log[OH  (aq)] pH + pOH = 14 Bases = [OH  (aq)] Acids = [H 3 O + (aq)],

[H 3 O + (aq)] = 10 -pH  you could also be given the pH or pOH and asked to calculate the [H 3 O + (aq)] or [OH - (aq)] [OH  (aq)]= 10 -pOH

Try These: 1. pH 4.0 [H 3 O + (aq)] = 2. pH 6.21 [H 3 O + (aq)] = 3. pH [H 3 O + (aq)] = 4. pH 7 [H 3 O + (aq)] = 5. pOH 1.0[OH  (aq)] = 6. pOH 13.2[OH  (aq)] = 7. pOH 6.90[OH  (aq)] = 8. pOH 0.786[OH  (aq)] = 0.1 mol/L 6  mol/L 1.3  mol/L mol/L 1 x mol/L 6.2 x mol/L mol/L 3.98 x mol/L

Summary of pH/pOH and concentration -log [H 3 O + ] = to find pH -log [OH - ] = to find OH 10 -pH = to find concentration of [H 3 O + ] 10 -pOH = to find concentration of [OH - ]

9. Complete the following table: [H 3 O + (aq)] [OH  (aq)] pHpOHAcid/Base/ Neutral 4.0 x mol/L 10 mol/L 2.0  10  11 mol/L x mol/L x mol/L 1.0 x mol/L 10 mol/L 1.0 x mol/L 3.16 x mol/L3.16 x mol/L 2.3 x mol/L mol/L acid base acid base

D. Measuring pH  pH can be measured using : 1. acid-base indicators 2. pH meter Indicators  an is any chemical that in an acidic or basic solution acid-base indicator changes colour  they can be dried onto strips of paper eg) litmus paper, pH paper

 they can be solutions eg) bromothymol blue, universal indicator, indigo carmine etc  they can be made from natural substances eg) tea, red cabbage juice, grape juice

 each indicator has a where it will specific pH range change colour  you can use to approximate the two or more indicators pH of a solution

pH Meters  using a pH meter is the most way of measuring  it has an that compares the [H 3 O + (aq)] in the solution to a and it will give a of the pH precise electrode standard digital readout pH

E. Diluting an Acid or Base  when you to an, you change the  diluting an acid will the until a pH of is reached [H 3 O + (aq)] or the [OH  (aq)] add wateracid or base decrease [H 3 O + (aq)] 7.0  diluting a base will the until a pH of is reached decrease [OH - (aq)] 7.0

Remember: C i V i = C f V f  A concentrated solution is made by dissolving 5g of HCl into 30 L of water. You then take 10 mL of this solution and dilute it to a volume of 50 L. What is the pH of the diluted solution?

Formulas to remember: pH = - log [H 3 O + ] pOH = - log [OH - ] [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH C = n/v C i V i = C f V f

 Review Assignment:  Textbook p. 244 #1-28