C5
- Created by: helloiamjessica
- Created on: 02-05-15 00:28
C5A - Moles and Molar Mass
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
C5A - Moles and Molar Mass
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
- WRITE OUT BALANCED SYMBOL EQUATION
C5B - Percentage Composition and Empirical Formula
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
C5B - Percentage Composition and 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
C5C - Quantitive Analysis
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
C5C - Quantitive Analysis
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
- WORK OUT RATIO - DIVIDE TWO CONCENTRATIONS TO GET NUMBER LESS THAN ONE
C5C - Quantitive Analysis
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
C5D - Titrations
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
C5D - Titrations
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
C5D - Titrations
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
C5D - Titrations
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
C5E - Gas Volumes
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
C5E - Gas Volumes
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
C5E - Gas Volumes
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
C5F - Equilibrium
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
C5F - Equilibrium
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
C5F - Equilibrium
- 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
C5F - Equilibrium
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
C5G - Strong and Weak Acids
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
C5G - Strong and Weak Acids
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
C5G - Strong and Weak Acids
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
C5H - Ionic Equations and Precipitation
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)
C5H - Ionic Equations and Precipitation
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)
C5H - Ionic Equations and Precipitation
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
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