C5

THE MOLE - 6.023 X 10 TO THE POWER OF 23

• WHEN YOU GET PRECISELY THAT NUMBER OF ATOMS OR MOLECULES OF ANY ELEMENT OR COMPOUND, THEY WEIGH EXACTLY THE SAME NUMBER OF GRAMS AS THEIR RELATIVE ATOMIC MASS
• ONE MOLE OF ATOMS OR MOLECULES OF ANY SUBSTANCE WILL HAVE A MASS IN GRAMS EQUAL TO THE RELATIVE FORMULA MASS FOR THAT SUBSTANCE
• E.G. CARBON - RELATIVE ATOMIC MASS OF 12 - ONE MOLE WEIGHS 12 G
• MOLAR MASS - THE MASS OF ONE MOLE - MEASURED IN G/MOL E.G. CARBON HAS MOLAR MASS OF 12G/MOL

NUMBER OF MOLES = MASS (G) OF ELEMENT OR COMPOUND / Mr OF ELEMENT OR COMPOUND

RELATIVE MASSES

• TOO TINY TO WEIGH - MASSES COMPARED WITH 1/12TH OF THE MASS OF CARBON-12 (ISOTOPE OF C)
• THE RELATIVE ATOMIC MASS OF AN ELEMENT IS THE AVERAGE MASS OF AN ATOM COMPAIRED TO THE MASS OF 1/12TH OF AN ATOM OF CARBON-12

CALCULATING MASSES IN REACTIONS USING MOLES

• E.G. CALCULATE MASS OF ALUMINIUM OXIDE FORMED WHEN 135G OF ALUMINIUM OXIDE IS BURNED IN AIR
  • WRITE OUT BALANCED SYMBOL EQUATION
    • 4Al + 3O2 ---> 2Al2O3
  • CALUCULATE NO. OF MOLES OF GIVEN ELEMENT/COMPOUND FOR GIVEN MASS USING FORMULA
    • MOLES = MASS / Mr = 135 / 27 = 5
  • LOOK AT RATIO OF MOLES IN EQUASION
    • 4 MOLES OF Al REACT TO PRODUCE 2 MOLES OF Al2O3 - HALF NUMBER OF MOLES ARE PRODUCED - 5 MOLES OF Al2 WILL REACT TO PRODUCE 2.5 MOLES OF Al2O3
  • CALCULATE THE MASS OF WORKED OUT NUMBER OF GIVEN SUBSTANCE
    • MASS = MOLES X Mr = 2.5 X 102 = 255G

CALCULATING PERCENTAGE COMPOSITION BY MASS OF COMPOUNDS

• E.G. 42.9G OF POTASSIUM IN 61.6G OF POTASSIUM HYDROXIDE:
  • USE EXPERIMENTAL DATA TO CALCULATE PERCENTAGE COMPOSITION OF K IN KOH: (42.9 / 61.6) X 100
  • OR USE THIS FORMULA:
• PERCENTAGE COMPOSITION BY MASS = Ar X NO. OF ATOMS OF ELEMENT / Mr OF WHOLE COMPOUND

EMPIRICAL FORMULAS - SIMPLIST RATIO OF ATOMS IN A COMPOUND

• LIST ALL ELEMENTS IN COMPOUND
• UNDERNEATH THEM, WRITE EXPERIMENTAL MASSES OR PERCENTAGES
• DIVIDE EACH MASS OR PERCENTAGE BY Ar OF PARTICULAR ELEMENT TO FIND NO. OF MOLES
• ROUND UP NUMBERS TO WHOLE NUMBERS IF NECESSARY AND TURN INTO SIMPLE RATIO BY DIVIDING BY WELL CHOSEN NUMBERS
• GET RATIO IN SIMPLEST FORM - TELLS YOU EMPIRICAL FORMULA

E.G. FIND EMPIRICAL FORMULA OF IRON OXIDE PRODUCED WHEN 44.8G OF IRON REACTS WITH 19.2 G OXYGEN

• LIST TWO ELEMENTS:                                             Fe (Ar 56)                                        O (Ar 16)
• WRITE EXPERIMENTAL MASSES                              44.8                                                 19.2
• DIVIDE BY Ar OF EACH ELEMENT                      44.8 / 56 = 0.8                                  19.2 / 16 = 1.2
• MULTIPLY BY 10...                                                       8                                                      12
• ...DIVIDE BY 4                                                               2                                                       3
• EMPIRICAL FORMULA = Fe2O3

CONCERNTRATION (G PER DM3) = NUMBER OF MOLES / VOLUME (DM3)

• THE MORE SOLUTE YOU DISSOLVE IN A GIVEN VOLUME, THE MORE CROWDED THE SOLUTE MOLECULES ARE AND THE MORE CONCENTRATED THE SOLUTION
• MEASURED IN GRAMS PER DECAMETER CUBED - NEED TO CONVERT UNITS GIVEN IN QUESTION:
  • 1 DM3 (1L) = 1 CM3 / 1000
• E.G. HOW MANY MOLES OF SODIUM CHLORIDE IN 250CM3 OF 3 M (3 MOLAR) SOLUTION?
  • CONVERT UNITS: 250 / 1000 = 0.25
  • REARRANGE EQUASION: MOLES = CONCERNTRATION X VOLUME = 3 X 0.25 = 0.75 MOLES

CONVERTING MOLES PER DM3 TO GRAMS PER DM3

• MAY BE ASKED TO ANSWER IN G PER DM3 INSTEAD OF MOLES PER DM3
  • WORK OUT RELATIVE FORMULA MASS FOR SOLUTE
  • CONVERT CONCENTRATION PER MOLES INTO CONCENTRATION PER GRAMS
  • MASS IN GRAMS = MOLES X RELATIVE FORMULA MASS

WORKING OUT DILUTION NEEDED

• MAY BE GIVEN CONCENTRATED SOLUTION AND ASKED TO DILUTE IT TO MAKE IT WEAKER:
  • WORK OUT RATIO OF TWO CONCENTRATIONS
  • MULTIPLY RATIO BY VOLUME OF SOLUTION YOU WANT TO END UP WITH (TELLS YOU HOW MUCH OF ORIGIONAL ACID NEED TO DILUTE)
  • WORK OUT VOLUME OF WATER NEEDED
• E.G. PRODUCE 500CM3 OF A 0.1 MOL/DM3 SOLUTION OF KOH IF GIVEN 1 MOL/DM3 SOLUTION AND WATER
  • WORK OUT RATIO - DIVIDE TWO CONCENTRATIONS TO GET NUMBER LESS THAN ONE
    • 0.1 / 1 = 1/10 - 1 TO 10 RATIO - 1 IN 10 DILUTION
  • DIVIDE BY WHAT YOU WANT TO END UP WITH
    • VOLUME OF DILUTE = RATIO X FINAL VOLUME = 1/10 X 500 = 60CM3
  • WORK OUT VOLUME NEEDED
    • VOLUME OF WATER = TOTAL VOLUME - VOLUME OF DILUTE = 500CM3 - 50CM3 = 450CM3

GUIDELINE DAILY AMOUNTS

• GUIDELINE DAILY AMOUNTS - THE AMOUNTS OF NUTRIENTS A HEALTHY ADULT SHOULD EAT IN A DAY IN A HEALTHY DIET
• PACKAGING - NUTRITIONAL INFORMATION LABELS - TELL YOU AMOUNTS OF NUTRIENTS IN FOOD - TELL YOU PERCENTAGE OF VARIOUS GDA'S PROCUT WILL SUPPLY
• AMOUNTS MAY NOT ALWAYS BE AMOUNT EATEN BECAUSE:
  • AMOUNTS GIVEN PER 100G OF FOOD - MAY EAT MORE OR LESS THAN THIS
  • MAY ADD OTHER THINGS E.G. CEREAL EATEN WITH MILK

ESTIMATING MASS OF SALT (SODIUM CHLORIDE) FROM SODIUM CONTENT

• E.G. BREAD CONTAINS 0.2G OF SODIUM (Ar23) - HOW MUCH SALT (SODIUM CHLORIDE - Mr 58.5)?
  • FIND RATIO OF SODIUM CHLORIDES Mr TO Ar: 58.5 / 23 = 2.543...
  • MULTIPLY BY AMOUNT OF SODIUM: 2.543 X 0.2 = 0.5086 G
• PROBABLY BE OVERESTIMATE - SODIUM WON'T ALWAYS COME FROM SODIUM CHLORIDE - OTHER COMPOUNDS TOO E.G. SODIUM NITRATE USED AS PRESERVATIVE

TITRATIONS - ALLOW TO FIND OUT EXACTLY HOW MUCH ACID NEEDED TO NEUTRILISE ALKILI (OR VICE VERSA)

• EQUIPMENT NEEDED:
  • PIPETTE - MEASURE ONE VOLUME OF SOLUTION
  • PIPETTE FILLER - STOPS INHALATION OF ALKALI
  • CONICAL FLASK - HOLDS MIXTURE
  • INDICATOR - TELLS YOU WHEN MIXTURE PH CHANGES
  • BURETTE - MEASURE VOLUMES - LET YOU ADD SOLUTION DROP BY DROP - KNOW QUANTITY USED
• USING PIPETTE AND PIPETTE FILLER, ADD KNOWN VOLUME (USUALLY 25CM3) ALKALI TO CONICAL FLASK ALONG WITH TWO/THREE DROPS OF INDICATOR
• FILL BURETTE WITH ACID BELOW EYE LEVEL TO AVOID GETTING IN EYE 
• ADD ACID TO ALKALI BIT AT A TIME - SWIRL FLASK REGULARLY - GO ESPECIALLY SLOWLY WHEN END POINT ABOUT TO BE REACHED
• END POINT - POINT WHERE COLOUR CHANGE HAPPENS - WHERE ACID/ALKILI IS NEUTRELISED
• RECORD VOLUME USED

WHY YOU NEED TO GET SEVERAL CONSISTANT READINGS

• TO INCREASE ACCURACY OF TITRATION - SPOT ANOMALOUS RESULTS
• FIRST TITRATION - ROUGH TITRATION - GET APPROXIMATE IDEA OF WHERE SOLUTION CHANGES COLOUR
• NEED TO REPEAT WHOLE THING A FEW TIMES - MAKE SURE GET ROUGHLY SAME RESULTS (WITHIN 0.2CM)

INDICATORS

• UNIVERSAL INDICATOR - USED TO ESTIMATE PH OF SOLUTION - TURNS VERIETY OF COLOURS - EACH REPRESENTS NARROW RANGE OF PH VALUES
• MADE OF MIXTURE OF DIFFERENT INDICATORS - COLOUR GRADUALLY CHANGES FROM RED (ACIDIC) TO PURPLE (ALKILINE))
• NOT USED IN TITRATIONS - YOU NEED TO SEE A SUDDEN COLOUR CHANGE AT THE END POINT
• NEED A SINGLE INDICATOR E.G. PHENOLPHTHALEIN - TURNS FROM CLEAR TO PINK IN PH'S HIGHER THAN 7

pH CURVES

• GRADUAL INCREASE IN PH AS ALKALI IS ADDED TO ACID
• END POINT - SUDDEN CHANGE (ALMOST VERITCAL LINE) - WHERE PH IS 7 - THIS GRAPH, 25CM3 ADDED
• VOLUME OF ALKALI NEEDED TO NEUTRILISE SOLUTION HERE IS 25CM3
• WHEN ADDING ACID TO ALKALI, CURVE WILL BE FLIPPED

CALCULATING CONCENTRATION

• METHOD:
  • WORK OUT HOW MANY MOLES OF KNOWN SUBSTANCE YOU HAVE (N = C X V)
  • WRITE DOWN BALANCED SYMBOL EQUASION FOR REACTION - WORK OUT HOW MANY MOLES OF UNKNOWN SUBSTANCE YOU MUST HAVE HAD
  • WORK OUT CONCENTRATION OF UNKNOWN STUFF (C = N/V)
• E.G. 25CM3 SODIUM HYDROXIDE IN FLASK OF CONCENTRATION 0.1 MOL/DM3 - TITRATION TAKES 49CM3 HYDROCHLORIC ACID TO NEUTRILISE - CONCENTRATION OF HYDROCHLORIC ACID?
  • MOLES OF KNOWN SUBSTANCE = 0.1 MOL/DM3 X (25/1000) DM3 = 0.0025 MOLS OF SODIUM HYDROXIDE
  • BALANCED SYMBOL EQUASION = NaOH + HCl ---> NaCl + H2O
  • MOLES OF UNKNOWN SUBSTANCE - USE EQUASION TO WORK OUT THAT, WITH EVERY MOLE OF KNOWN SUBSTANCE, YOU MUST HAVE HAD ABOUT 1 MOLE OF HYDROCHLORIC ACID = 0.0025MOLS NaOH
  • CONCENTRATION = 0.0025 MOL / (49/1000) DM3 = 0.0510 MOL/DM3

COLLECTION METHODS

• TWO METHODS - BOTH USE CONICAL FLASKS AS STANDARD APERATUS TO COLLECT GASES - WHAT YOU CONNECT DEPENDS ON WHAT TRYING TO COLLECT:
  • GAS SYRINGE - USED FOR COLLECTING ANY GAS - USUALLY GIVE VOLUME ACCURATE TO NEAREST CM3 - IF REACTION TOO VIGOROUS, PLUNGER CAN BE BLOWN OUT OF END OF SYRINGE
  • 
  • UPTURNED MEASURING CYLINDER OR BURETTE - USE DELIVERY TUBE TO BUBBLE GAS INTO UPSIDE DOWN MEASURING CYLINDER OR GAS JAR FILLED WITH WATER - NO GOOD FOR COLLECTING THINGS WHICH DISSOLVE IN WATER E.G. AMMONIA, HYDROGEN CHLORIDE - UPTURNED BURETTE IS MORE ACCURATE - MEASURES TO NEAREST 0.1CM3
  • 

MEASURING MASS OF GAS PRODUCED

• MEASURE MASS OF GAS PRODUCED BY CARRYING OUT EXPERIMENT ON A MASS BALANCE
• GAS RELEASED - AMOUNT GONE EASILY MEASURED BY MASS BALANCE
• MOST ACCURATE OF THREE METHODS - MASS BALANCE VERY ACCURATE
• DISADVANTAGE - RELEASES GAS SRAIGHT INTO ROOM

ONE MOLE OF ANY GAS ALWAYS OCCUPIES 24DM3 AT ROOM TEMPERATURE AND PRESSURE (RTP - 25 DEGREES AND ONE ATMOSPHERES)

STOPPING OF REACTIONS - WHEN ONE REACTANT IS USED UP

• REACTION STOPS WHEN ALL REACTANT IS USED UP - ANY OTHER REACTANTS ARE EXCESS
• LIMITING REACTANT - REACTANT USED UP
• AMOUNT OF PRODUCT FORMED DIRECTLY PROPORTIONAL TO LIMITING REACTANT - E.G. IF LIMITING REACTANT DOUBLED, YEILD WILL DOUBLE
• REASON - IF YOU ADD MORE REACTANT THERE WILL BE MORE REACTANT PARTICLES TO TAKE PART IN REACTION - MORE PRODUCT PARTICLES

READING GRAPHS AND TABLES OF REACTIONS

• GRAPH GOES FLAT - Y AXIS - TOTAL VOLUME OF GAS PRODUCED 
  • X AXIS - TOTAL TIME OF REACTION
• STEEPER CURVE - REACTION FAST
• CURVE STARTS GETTING LESS STEEP - REACTION SLOWING DOWN - GETTING READY TO STOP
• MORE THAN ONE CURVE ON GRAPH:
  • STEEPEST CURVE WILL BE QUICKEST REACTION
  • CURVE THAT GOES HORIZONTAL AT HIGHEST POINT PRODUCES MOST PRODUCT - LEAST AMOUNT OF LIMITING REACTANT
  • ALL CURVES GO FLAT AT SAME HEIGHT - SAME AMOUNT OF LIMITING REACTANTS

REVERSABLE REACTIONS - REACTIONS WHERE THE PRODUCTS OF THE REACTION CAN THEMSELVES REACT TO                                                    GIVE THE ORIGIONAL REACTANTS - A + B <---> C + D

REACHING EQUILIBRIUM - FORWARD REACTION IS GOING AT EXACT SAME RATE AS BACKWARD REACTION

• AS REACTANTS A AND B REACT, THEIR CONCENTRATIONS FALL - FORWARD REACTION WILL SLOW DOWN
• AS MORE AND MORE PRODUCTS (C AND D) ARE FORMED, THEIR CONCENTRATIONS WILL RISE - BACKWARD REACTION WILL SPEED UP
• AFTER WHILE, FORWARD AND BACKWARD REACTION WILL BE GOING AT SAME RATE - EQUILIBRIUM REACHED
• BOTH REACTIONS STILL HAPPENING BUT THERE IS NO OVERALL EFFECT - CONCENTRATIONS OF REACTANTS AND PRODUCTS WILL HAVE REACHED A BALANCE AND WON'T CHANGE
• ONLY REACHED IF REVERSABLE REACTION TAKES PLACED IN A CLOSED SYSTEM - NONE OF REACTANTS OR PRODUCTS CAN ESCAPE

POSITION OF EQUILIBRIUM

• EQUILIBRIUM LIES TO RIGHT - CONCENTRATION OF PRODUCT GREATER THAN CONCENTRATION OF REACTANT
• EQUILIBRIUM LIES TO LEFT - CONCENTRATION OF REACTANT GREATER THAN CONCENTRATION OF PRODUCT

FACTORS WHICH CHANGE POSITION OF EQUILIBRIUM

• TEMPERATURE 
• PRESSURE (ONLY EFFECTS GASES)
• CONCENTRATION

ADDING A CATALYST DOES NOT CHANGE EQUILIBRIUM POSITION - SPEED UP FORWARD AND BACKWARDS REACTION BY THE SAME AMOUNT - REACTION REACHES EQ. QUICKER BUT SAME AMOUNT PRODUCT FORMED

EQUILIBRIUM COUNTERACTS ANY CHANGE MADE

• TEMPERATURE - ALL REACTIONS ARE EXOTHERMIC IN ONE DIRECTION AND ENDOTHERMIC IN ANOTHER
  • TEMP DECREASED - EQUILIBRIUM MOVE TO INCREASE IT - EQUILIBRIUM MOVES IN ENDOTHERMIC DIRECTION TO PRODUCE MORE HEAT
  • RAISE TEMPERATURE - EQUILIBRIUM MOVE IN ENDOTHERMIC DIRECTION TO DECREASE IT
• PRESSURE - ONLY AFFECTS EQUILIBRIUM INVOLVING GASES
  • PRESSURE INCREASED - EQUILIBRIUM MOVES IN DIRECTION WHERE FEWER MOLES OF GAS
  • PRESSURE DECREASED - EQUILIBRIUM MOVES IN DIRECTION WHERE MORE MOLES OF GAS

• CONCENTRATION - 
  • INCREASE CONCENTRATION OF REACTANTS - EQUILIBRIUM DECREASES IT BY SHIFTING TO THE RIGHT (MAKE MORE PRODUCT)
  • INCREASE CONCENTRATION OF PRODUCT - EQUILIBRIUM MOVES TO LEFT - MAKES MORE REACTANTS

EQUILIBRIUM TABLES AND GRAPHS

• AS PRESSURE INCREASES, PROPORTION OF AMMONIA INCREASES - INCREASED PRESSURE SHIFTS EQUILIBRIUM TO SIDE WITH FEWER MOLES OF GAS
• MULTIPLE LINES ON GRAPHS:
  • EACH LINE REPRESENTS A DIFFERENT TEMP
  • TEMP INCREASES - PROPORTION OF AMMONIA DECREASES (BACKWARDS REACTION ENDOTHERMIC - EQUILIBRIUM TRYING TO CHANGE TO REDUCE TEMP)
  • CONDITIONS WHICH GIVE MOST PRODUCT - HIGHEST CURVE ON GRAPH

THE CONTACT PROCESS - THREE STAGES:

• FIRST STAGE - BURN SULFUR IN AIR TO MAKE SULFUR DIOXIDE - S (s) + O2 (g) ---> SO2 (g)
• SECOND STAGE - SULPHUR DIOXIDE OXIDISED WITH CATALYST TO MAKE SULPHUR TRIOXIDE
  • 2SO2 (g) + O2 (g) <---> 2SO2 (g)
• SULFUR TRIOXIDE USED TO MAKE SULPHURIC ACID - SO2 (g) + H2O (l) ---> H2SO4 (aq)

CONDITIONS USED FOR SECOND STAGE 

• 2SO2 +O2 <---> 2SO3 - REVERSABLE REACTION
• TEMPERATURE - MAKING SULPHUR TRIOXIDE IS AN EXOTHERMIC REACTION
  • TEMP NEEDS TO BE REDUCED TO SHIFT EQUILIBRIUM TO RIGHT TO GET MORE PRODUCT
  • REDUCING TEMP SLOWS REACTION - COMPROMISE TEMP OF 450 DEGREES USED - YEILD AND SPEED
• PRESSURE - TWO MOLES OF PRODUCT COMPARED TO 3 MOLES OF REACTANT
  • MORE  PRODUCT - PRESSURE INCREASED - EQUIIBRIUM SHIFTS TO RIGHT TO REDUCE PRESSURE
  • INCREASING PRESSURE EXPENSIVE - NOT NECESSARY - EQUILIBRIUM ALREADY ON RIGHT
  • ATMOSPHERIC PRESSURE (1 ATMS) USED
• CALTALYST - VANADIUM PETOXIDE (V2O5) - INCREASE RATE OF REACTION - DOESN'T AFFECT EQUI. POSITION 

ACIDS - STRONG OR WEAK?

• ACIDS IONISE IN WATER - PRODUCE H+ IONS - E.G. HNO3 ---> H+ + NO3-
  • STRONG - IONISE COMPLETELY IN WATER - LOTS OF H+ RELEASED E.G. HYDROCHLORIC, SULPHURIC
  • WEAK - DON'T FULLY IONISE - SMALL NUMBERS OF H+ RELEASED E.G. ETHANOIC, CITRIC, CARBONIC
• IONISATION OF WEAK ACID - REVERSABLE REACTION - SETS UP EQUILIBRIUM MIXTURE - ONLY FEW H+ IONS RELEASED - EQUILIBRIUM LIES TO LEFT
• PH - MEASURE OF CONCENTRATION OF H+ IONS IN SOLUTION - STRONG HAVE PH 1 OR 2 ETC.
• EXAMPLES OF STRONG AND WEAK ACIDS:
  • STRONG - HCl ---> H+ + Cl-
  • WEAK - CH3COOH <---> H+ + CH3COO-

CONCENTRATION AND STRENGTH

• STRENGTH - DESCRIBES PROPORTION OF ACID MOLECULES IONISE IN WATER
• CONCENTRATION - MEASURES HOW MANY MOLES OF ACID THERE ARE IN A DM3 OF WATER - HOW WATERED
  • DESCRIBES TOTAL NUMBER OF DISSOLVED ACID MOLECULES - NOT NUMBER THAT PRODUCE H IONS
  • MORE ACID MOLECULES PER DM3 = MORE CONCENTRATED - CAN HAVE CONCENTRATED WEAK ACID

REACTION RATES

• HYDROCHLORIC ACID (STRONG) AND ETHENOIC ACID (WEAK) BOTH REACT WITH Mg TO GIVE HYROGEN
  • 2HCl + Mg ---> MgCl2 + H2
  • 2CH3COOH + Mg ---> Mg(CH3COO)2 + H2
• HYDROCHLORIC ACID AND ETHANNOIC ACID REACT WITH CALCIUM CARBONATE TO GIVE CO2
  • 2HCL + 2CaCO3 ---> CaCl2 + H2O + CO2
  • 2CH3COOH + CaCO3 ---> Ca(CH3COO)2 + H2O + CO2
• DIFFERENCE - RATE OF REACTION - HCL REACTS FASTER THAN ETHANOIC OF SAME CONCENTRATION  - HCL STRONG ACID AND ETHANOIC WEAK
• WHEN WEAK ACID PUT INTO WATER, RELEASES FEW H+ IONS - CONCENTRATION OF H+ IONS LOW COMPARED TO STRONG ACID - WHEN MAGNESIUM ADDED, COLLISION FREQUENCY LOWER
• H+ IONS REACT - CONCENTRATION OF H+ DECREASES - EQUILIBRIUM SHIFTS TO COMPENSATE - MORE H+ IONS RELEASED - THESE IONS REACT - EQUILIBRIUM SHFTS AND SO ON - AS IONS REMOVED, MORE SUPPLIED
• DIFFERENT TO STRONG ACID - ALL PARTICLES IONISED - LOADS OF H+ IONS RELEASED - WHEN MG ADDED, COLLISION FREQUENCY HIGH                                                                                                                                                                                                                                                                                                                                     

GAS VOLUMES

• HYDROCHLORIC ACID WILL REACT FASTER THAN ETHANOIC BUT AMOUNT OF YEILD WILL BE THE SAME
• IF CONCENTRATIONS SAME, NUMBER OF MOLECULES IN A LITRE OF WATER WILL BE SAME
• EACH OF THESE MOLECULES CAN LET GO OF ONE HYDROGEN ION
  • HCL ---> H+ + CL-
  • CH3COOH <---> H+ + CH3COO-
• HYDROCHLORIC ACID HAS LET GO OF THEM ALL AT ONCE, RATHER THAN GRADUALLY 
• SINCE TOTAL NUMBER OF H+ IONS AVAILABLE IS SAME, VOLUME OF GASEOUS PRODUCTS SAME

CONDUCTIVITY

• ETHANOIC ACID - LOWER ELECTRICAL CONDUCTIVITY THAN SAME CONCENTRATION OF HYDROCHLORIC
• IONS CARRY CHARGE THROUGH ACID SOLUTIONS AS THEY MOVE - LOWER CONCENTRATION OF IONS MEANS LESS CHARGE CAN BE CARRIED
• ELECTROLYSIS OF HCL AND ETHANOIC - PRODUCES H2 - BOTH PRODUCE H+ IONS

PRECIPITATION REACTIONS - TWO SOLUTIONS REACTING TOGETHER TO MAKE AN INSOLUBLE SUBSTANCE

• PRECIPITATE - INSOLUBLE SUBSTANCE FORMED FROM PRECIPITATION REACTION - MAKES SOLUTION CLOUDY
• MOST PRECIPITATION REACTIONS IONS - HAVE TO BE ABLE TO MOVE - COLLIDE WITH EACHOTHER AND REACT - IONIC SUBSTANCES HAVE TO BE IN SOLUTION OR MOLTEN TO REACT
• USUALLY EXTREMELY QUICK - HIGH COLLISION FREQUENCY BETWEEN IONS

IONIC EQUATIONS

• BARIUM CHLORIDE + SODIUM SULPHATE ---> BARIUM SULPHATE + SODIUM CHLORIDE
•         BaCl2 (aq)          +         NaSO4 (aq)        --->     BaSO4 (s)               +      2NaCl (aq)
• PRECIPITATION REACTION - START WITH TWO SOLUTIONS AND END UP WITH SOLID
• BARIUM AND SULPHATE IONS FORM PRECIPITATE
• SPECTATOR IONS - IONS THAT DON'T CHANGE DURING REACTION
• SODIUM AND CHLORIDE IONS - DISSOLVED IN SOLUTION BEFORE REACTION- STILL DISSOLVED AFTERWARDS - SPECTATOR IONS
• IONIC EQUATION - CONCENTRATES ON PARTS OF REACTION THAT RACTS - IGNORES SPECTATOR IONS
• Ba 2+ (aq) + SO4 2- (aq) ---> BaSO4 (s) 

TESTING FOR SULFATE IONS (SO4 2-)

• ADD DILUTE HYDROCHLORIC ACID FOLLOWED BY BARIUM CHLORIDE
• WHITE PRECIPITATE OF BARIUM SULPHATE - MEANS ORIGIONAL COMPOUND WAS SULFATE
• E.G. ADDING HCL TO BARIUM CHLORIDE AND POTASSIUM SULFATE PRODUCES WHITE PRECIPITATE

TESTING FOR CHLORIDE (Cl-), BROMIDE (Br-) AND IODIDE (I-) IONS

• ADD DILUTE NITRIC ACID FOLLOWED BY LEAD NITRATE
  • CHLORIDE GIVES WHITE PRECIPITATE                                               Pb 2+ (aq) + 2Cl- (aq) ---> PbCl2 (s)
  • BROMIDE GIVES CREAM PRECIPITATE                                               Pb 2+ (aq) + 2Br- (aq) ---> PbBr2 (s)
  • IODIDE GIVES YELLOW PRECIPITATE                                                  Pb 2+ (aq) + 2I- (aq) ---> PbI2 (s)

PICKING REACTANTS AND SPECTATOR IONS

• TO MAKE INSOLUBLE SALT, IONS NEEDED - E.G. LEAD IODIDE NEEDS LEAD AND IODINE IONS - NEED TO BE IN SOLUTION TO MOVE AROUND
• NITRATES SOLUBLE - GET LEAD FROM LEAD NITRATE SOLUTION AND IODINE FROM POTASSIUM IODIDE
• Pb(NO3)2 (aq) + 2KI (aq) ---> PbI2 (s) + 2KNO3 (aq)                                    Pb 2+ (aq) + 2I- (aq) --> PbI2 (s)

PREPARING INSOLUBLE SALTS - METHOD

• STEP ONE - MIX SOLUTIONS OF REACTANTS
  • ADD 1 SPATULA OF LEAD NITRATE TO TEST TUBE, FILL WOTH DISTILLED WATER
  • SHAKE THOROUGHLY TO ENSURE LEAD NITRATE DISSOLVED
  • DO THE SAME WITH ONE SPATULA OF POTASSIUM IODIDE
  • TIP TWO SOLUTIONS INTO BEAKER, STIR AND SEE SALT PRECIPITATE OUT
• STEP TWO - FILTERATION
  • PUT FOLDED FILTER PAPER INTO FILTER FUNNEL - STICK FUNNEL INTO CONICAL FLASK
  • POUR CONTENTS OF BEAKER INTO MIDDLE OF FILTER PAPER, MAKING SURE NONE SPILLS DOWN SIDES OF FILTER PAPER SO SOLIDES DONT GET THROUGH
  • SWILL OUT BEAKER WITH MORE DISTILLED WATER - MAKE SURE ALL PRODUCT GOT FROM BEAKER
• STEP THREE - WASH AND DRY RESIDUE
  • RINSE FILTER PAPER WITH DISTILLED WATER - MAKE SURE ALL SOLUBLE SALTS WASHED AWAY
  • SCRAPE LEAD IODIDE ONTO FRESH FILTER PAPER AND LEAVE TO DRY