CHEM LAB Notes PDF 1st Sem 2024-2025

CHEM LAB 1ST SEM | A.Y. 2024-2025 EXPERIMENT 1 | COMMON LABORATORY OPERATIONS ➔ Evaporation - the process of turning a liquid into a vapor or gas A. To become acquainted with some by heating it, causin...

CHEM LAB 1ST SEM | A.Y. 2024-2025 EXPERIMENT 1 | COMMON LABORATORY OPERATIONS ➔ Evaporation - the process of turning a liquid into a vapor or gas A. To become acquainted with some by heating it, causing the liquid to common laboratory operations gradually disappear and leave behind any dissolved solids. EXPERIMENT/UNIT TOPIC OUTLINE I. MEASURING AND ESTIMATING I. MEASURING AND ESTIMATING LIQUID VOLUMES LIQUID VOLUMES II. TRANSFERRING LIQUIDS III. OPERATING BUNSEN BURNER IV. HEATING LIQUIDS IN TEST TUBES V. PRECIPITATION APPARATUS USED VI. FILTRATION VII. DECANTATION ➔ Graduated Cylinder VIII. EVAPORATION ➔ Test Tubes ➔ Test Tube Holder Terminology MATERIALS USED ➔ Tap Water ★ Terms Used PROCEDURE 1 ➔ Meniscus - The curved upper surface of a liquid in a tube. Fill your graduated cylinder to the 10 ml ➔ Supernatant Liquid- the liquid that Read the volume at eye level, lies above a sediment of aligning with the lower meniscus precipitate Transfer the 2 ml of water to the calibrated test tubes and describe ➔ Residue- The solid resulting its level Gather 3 non-calibrated test tubes ➔ Precipitation - Is the process where Transferring the 2 ml of water to the a solid forms and settles out of a non calibrated test tube liquid mixture during a chemical reaction. ➔ Filtration - Used to separate solid ★ Important Note from liquid by passing the mixture through a filter. ➔ When measuring transparent ➔ Decantation - the process of liquid (ex. water) use the Lower separating a liquid from a solid or meniscus to read the volume another liquid by carefully pouring off the top layer without disturbing ➔ When measuring dark colored the sediment or bottom layer. liquid use the upper meniscus to read the volume  OBSERVATIONS FOR PROCEDURE 1 III. OPERATING BUNSEN BURNER The 2 mL of water filled approximately 1/5 APPARATUS USED of the non-calibrated test tube, while 5 ➔ Bunsen Burner mL filled about half of it, and 10 mL nearly reached the top. This visual reference MATERIALS USED provides a useful guide for estimating ➔ Torch Lighter liquid volumes in non-calibrated PROCEDURE 1 glassware. Variations in test tube size may slightly affect these estimates. ➔ important to ensure that all the components are in their proper positions II. TRANSFERRING LIQUIDS ➔ the gas outlet, gas adjustment knob, and air holes should be APPARATUS USED closed. ➔ Beakers PROCEDURE 2 ➔ Erlenmeyer Flask ➔ When ready, open the gas valve ➔ Glass rod first, followed by adjusting the gas PROCEDURE 1 knob halfway. ➔ One beaker with ¾ full of water ➔ Ignite the burner using a torch ➔ Glass rod was used to pour the lighter. You can then control the water into the empty vessel flame size by adjusting the gas knob QUESTIONS FOR PROCEDURE 2 ➔ Gradually open the air holes until you achieve the desired blue flame. ➔ Why should one use a glass rod or a piece of glass tubing in QUESTIONS FOR PROCEDURE 2 transferring liquids from one vessel ➔ Why is it important to achieve a to another? blue flame on a Bunsen burner? Answer: using a glass rod to transfer Answer: A blue flame indicates efficient liquid ensures a controlled and combustion with proper air flow, making smoothflow, reducing the risk of spills or it hotter and more suitable for heating in splashing. experiments. ➔ What are luminous and non luminous flames? ➔ Answer: Non luminous flame example is a blue flame because it  indicates complete combustion. The luminous flame example is red flame because it indicates incomplete combustion. III. OPERATING BUNSEN BURNER V. PRECIPITATION IV. HEATING LIQUIDS IN TEST TUBES APPARATUS USED APPARATUS USED ➔ Test tubes ➔ Test tubes MATERIALS USED ➔ Test Tube holder MATERIALS USED ➔ 1 mL Ferric Chloride Solution ➔ 4 mL Sodium Hydroxide Solution ➔ Tap Water PROCEDURE 1 PROCEDURE 1 ➔ Add 1 mL ferric chloride solution ➔ 1 test tube with half filled water and pour it into a test tube. ➔ In a 45° degree angle hold the test PROCEDURE 2 tube using the test tube holder ➔ Slowly move it back and forth ➔ Slowly add 4 mL of sodium ➔ Don’t heat the empty part of the hydroxide solution to the test tube. tube ➔ Observe the formation of solid hydroxide precipitate QUESTIONS FOR PROCEDURE 1 ➔ Allow the precipitate to settle to the ➔ Why should you heat the upper bottom of the test tube. portion of the liquid rather than the bottom QUESTIONS FOR PROCEDURE 2 Answer: heating the upper portion ➔ Why does ferric hydroxide form a prevents rapid boiling and splashing, precipitate while sodium chloride allowing liquid to heat evenly and safely remains in solution? ➔ What might happen if you heat the bottom of the test tube directly? Answer: Ferric Hydroxide doesn’t dissolve Answer: Direct heating of the bottom can in water, so it forms a solid (precipitate) cause rapid boiling, leading to sudden that settles at the bottom. Sodium splashing or potentially dangerous Chloride dissolves easily in water, so it "bumping" of the liquid. stays in the liquid. ➔ What is the chemical reaction equation for the formation of ferric hydroxide and sodium chloride? Answer: The reaction is FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl PROCEDURE 1 PROCEDURE 3 ➔ Fold filter paper in half ➔ Fold into quarters with top section ➔ Add a few more drops of sodium smaller than the bottom hydroxide solution without ➔ Tear off corner of smaller section agitating the test tube. ➔ Form an open cone ➔ Check if more precipitate forms. PROCEDURE 2 ➔ If more precipitate forms, allow it to settle. ➔ Place the prepared cone in the ➔ Continue adding drops of sodium funnel and moisten it with water to hydroxide until no more precipitate make it stick to the side of the forms. funnel. ➔ Suspend the funnel into you QUESTIONS FOR PROCEDURE 3 Erlenmeyer Flask. ➔ What observations would indicate that the precipitation reaction is QUESTIONS FOR PROCEDURE 2 complete? ➔ What is the purpose of moistening Answer: The reaction is complete when the filter paper before placing it in no more solid ferric hydroxide forms after the funnel? adding more sodium hydroxide. The Answer: Moisten the filter paper to help it solution should also be clear of stick to the funnel, preventing it from additional precipitate. slipping and ensuring a better seal, which helps avoid leakage of filtrate VI. PRECIPITATION ➔ What might happen if the filtrate paper is not properly placed in the APPARATUS USED funnel? Answer: If the filter paper isn’t properly ➔ Funnel placed, it might not catch all the solid ➔ Erlenmeyer Flask particles, leading to contamination of the ➔ Test tube you used in V. filtrate with residue or leakage. Precipitation VII. FILTRATION ➔ Filter Paper MATERIALS USED ➔ Transfer the content of the test tube ➔ Tap water to moisten the prepared used in V. Precipitation to the filter paper. cone ➔ This is Filtration  ➔ The liquid passes through the filter QUESTIONS FOR PROCEDURE 1 is known as filtrate ➔ What is the purpose of adding water to the precipitate before QUESTIONS FOR PROCEDURE 3 decanting? ➔ What is the difference between the Answer: Adding water helps to rinse and filtrate and residue? loosen any remaining particles of the Answer: The filtrate is the clear liquid that precipitate, making it easier to separate passes through the filtrate paper, while from the liquid the residue is the solid that remains on the filter paper. PROCEDURE 2 ➔ Shake the precipitate and water ➔ How can you ensure that the well → to mix thoroughly filtration process is complete and ➔ Allow the precipitate to settle → at effective? the bottom of the test tube Answer: Ensure all liquid has passed through the filter paper and that no solid particles remain in the filtrate. Check the QUESTIONS FOR PROCEDURE 2 residue on the filter paper to confirm that ➔ Why is it important to allow the no additional solids are left in the liquid. precipitate to settle before decanting? VII. DECANTATION Answer: It ensures that the solid particles have enough time to sink to the bottom, APPARATUS USED making it easier to pour off the clear liquid without disturbing the sediment. ➔ Test tube MATERIALS USED PROCEDURE 3 ➔ Tap water ➔ Slowly pour off the supernatant ➔ Precipitate liquid → being careful to keep the PROCEDURE 1 precipitate in the test tube ➔ Show evidence of decantation → to ➔ Transfer the precipitate from the your instructor filter paper → into a test tube ➔ Add 5 mL of water to the precipitate → in the test tube QUESTIONS FOR PROCEDURE 3 ➔ What observations indicate that decantation has been successful? Answer: Success is indicated by the clear Answer: Prevents the risk of burning or liquid you’ve poured off and the presence decomposing salt, which could affect the of a solid residue at the bottom of the result and make crystals impure or container change their appearance. ➔ What are the limitations of ➔ What is the purpose of using a clay decantation compared to filtration? flame shield and wire gauze during Answer: Decantation might not remove the process? fine particles that are suspended in the Answer: Both of them help distribute the liquid, whereas filtration can capture all heat evenly and protect the evaporating solid particles, including smaller ones dish from direct contact with the flame. that haven’t settled. PROCEDURE 2 VIII. EVAPORATION ➔ Allow the liquid to evaporate slowly APPARATUS USED → until all the liquid has evaporated ➔ Be careful not to overheat the ➔ Evaporating Dish crystals → formed during ➔ Bunsen Burner evaporation ➔ Clay Flame Shield ➔ Allow the contents to cool → the ➔ Wire Gauze remaining solid is called residue MATERIALS USED QUESTIONS FOR PROCEDURE 2 ➔ Sodium Chloride (5mL) ➔ Residue (Formed after evaporation) PROCEDURE 1 ➔ What should you do if you notice that the residue is not forming ➔ Transfer 5 mL of sodium chloride properly or seems discolored? solution → to an evaporating dish Answer: Check if the heating is too ➔ Connect and light the Bunsen intense or if the solution was burner → adjust the flame for contaminated. Adjust the heating and gentle heating ensure the process is carried out ➔ Place a clay flame shield over the carefully. burner → with wire gauze on top ➔ How can you ensure that all the liquid has been completely QUESTIONS FOR PROCEDURE 1 evaporated before cooling the residue? ➔ Why is it important to heat the Answer: Continue heating until no more solution gently during Evaporation? steam is visible and the result looks dry. You can also observe the residue to ensure no more liquid is left. ➔ Polar Substance - molecules that have an uneven distribution of electrical charge due to EXPERIMENT 2 | Solutions, Acid, and Bases differences in electronegativity B. To compare the solubility of a substance between the atoms involved. in various solvent and to study the ➔ Nonpolar substances - substances different types of solutions are molecules in which the electrons are distributed more evenly, resulting in no significant regions of partial positive or EXPERIMENT/ UNIT TOPIC OUTLINE negative charge. I. SOLUBILITY OF SOLIDS IN LIQUIDS ➔ Miscible - ability of two substances to mix together in all proportions, II. SOLUBILITY OF LIQUIDS IN LIQUIDS forming a homogeneous solution III. UNSATURATED, SATURATED, AND SUPERSATURATED IV. EXOTHERMIC AND ENDOTHERMIC I. SOLUBILITY OF SOLIDS IN LIQUIDS REACTIONS APPARATUS USED V. PROPERTIES OF ACID AND BASES VI. RELATIVE ACTIVITY OF ACIDS ➔ Test tubes VII. RELATIVE ACTIVITY OF BASES MATERIALS USED ➔ Iodine, Sand, Sugar, and Salt (solute) Terminology ➔ 20mL of water (solvent) ★ Terms Used ➔ 1mL of chloroform or carbon tetrachloride ➔ Solute - The substance being dissolved by the solvent (sand, salt, etc.) ★ Important Note ➔ Solvent - substance dissolving the solute (water) ➔ For the experiment 2, we only added the chloroform as the ➔ Solubility - is the ability of a carbon tetrachloride was substance to dissolve in a solvent unavailable to form a homogeneous solution at a specific temperature and pressure.  Answer: No, because solutes, which are solids being dissolved, have different PROCEDURE 1 rates of solubility. ➔ Small amounts of the solutes were ➔ In which solvent does iodine placed in separate test tubes. dissolve faster? ➔ 20mL of water (solvent) was added Answer: Iodine dissolves faster in to each Chloroform than in water because iodine and chloroform are both nonpolar ➔ The test tube was shaken to dissolve substances. the solute ➔ The results were recorded ★ Important Note THE SOLUBILITY OF THE SOLUTES ➔ Iodine (I₂) and chloroform (CHCl₃) are both nonpolar substances, ➔ IODINE - Slightly Soluble which is why iodine dissolves better in chloroform than in water. The ➔ TABLE SALT - Soluble solubility principle of "like ➔ SAND - Insoluble dissolves like" applies here: nonpolar solutes dissolve more ➔ SUGAR - Soluble easily in nonpolar solvents. ★ Important Note ➔ The iodine is slightly soluble. It II. SOLUBILITY OF LIQUIDS IN LIQUIDS takes time to dissolve especially if the iodine used is still compact and APPARATUS USED not the crushed iodine. ➔ Test tubes PROCEDURE 2 MATERIALS USED ➔ To the tube containing Iodine, 1mL of ➔ 2mL of ethyl alcohol (added chloroform was added. separately by 1mL each) ➔ The solubility was recorded in ➔ Water comparison to its solubility with ➔ Oil (coconut oil) water. ➔ Glycerine PROCEDURE 1 QUESTIONS FOR PROCEDURE 2 1mL of ethyl alcohol was poured ➔ Are all solids equally soluble in the into a test tube with an equal same solvent? volume of water inside. The test tube was shaken  MATERIALS USED QUESTIONS FOR PROCEDURE 1 Water Do alcohol and water mix in all Crystal Sodium Thiosulfate proportions? PROCEDURES Answer: Water and alcohol mix in all proportions because both substances Take 1mL of water and add a drop are polar and can form hydrogen bonds of a small crystal of sodium with each other. thiosulfate and shake the mixture What term is given to such liquids? Solution produced: Unsaturated solution Answer: Miscible To the solution in ( 1 ) add some more crystals of sodium thiosulfate until no more dissolves PROCEDURE 2 In two separate test tubes with 5mL Solution produced: Saturated solution of water, 1mL of glycerine and oil was added to each. (glycerine on Add more crystals to the solution in one test tube and oil to the other) (2) and heat the mixture (use The test tubes were shaken and the bunsen burner) solubilities of each in water were compared Solution produced: Supersaturated solution OBSERVATIONS PROCEDURE 2 Set aside the resulting solution to cool. Once the solution is cold, drop Glycerine dissolves completely in 1 crystal of sodium thiosulfate water because it is polar like water. In contrast, oil is nonpolar, which is Formed from cooling: Crystalized sodium why it is immiscible in water, thiosulfate leading to a distinct separation or layer. Solution remained: Saturated solution III. UNSATURATED, SATURATED, AND ★ Important Note SUPERSATURATED SOLUTIONS ➔ Saturated Solution - Contains the APPARATUS USED maximum amount of solute that can dissolve at a given ➔ Test tubes temperature. No more solute can ➔ Bunsen burner dissolve. (there are still crystals at ➔ Test tube clamp the bottom that cannot be dissolved)  ➔ Unsaturated Solution - Contains PROCEDURE 1 less solute than the maximum amount it can hold. More solute Place 10mL of water in the test tube can still dissolve. (appears clear) and note the temperature (no need for a thermometer). ➔ Supersaturated Solution - Contains Add solid sodium hydroxide and more solute than a saturated shake the tube solution at the same temperature. It's unstable and can crystallize Does it become hot or cold? easily. (with the help of heating, the Answer: HOT additional solute from the saturated was dissolved to add it to the solution and letting it have a Is it endothermic or exothermic? clear appearance even if the ratio Answer: Exothermic (heat releasing) of the solute is greater than the solvent) PROCEDURE 2 Difference of saturated and supersaturated Place 10mL of water in the test tube and note the temperature (no need Saturated: Holds the maximum for a thermometer). solute that can dissolve at a Add solid ammonium chloride to given temperature. the water and shake the tube. Supersaturated: Holds more Does it become hot or cold? solute than normally possible at Answer: COLD the same temperature, making it unstable. Is it endothermic or exothermic? Answer: Endothermic (heat absorbing) ★ Important Note IV. EXOTHERMIC AND ENDOTHERMIC REACTIONS ➔ Exothermic - heat releasing making the substance hot APPARATUS USED ➔ Endothermic - heat absorbing ➔ Test tubes making the substance cooler MATERIALS USED 20mL of water (separately added PROPERTIES OF ACID AND BASES by 10mL each) Solid sodium hydroxide Solid ammonium chloride APPARATUS USED ➔ Test tubes  Add a drop of methyl orange MATERIALS USED indicator to all the acids above. 1 mL of acetic acid Observation: Methyl orange indicator 1 mL of hydrochloric acid turned into red in the presence of acids, 1 mL of ammonium hydroxide showing their acidic nature. 1 mL of sodium hydroxide 4 mL of distilled water (equally Add a drop of phenolphthalein distributed to 4 separate test indicator to all the bases above. tubes) Methyl orange indicator Observation: Phenolphthalein indicator Phenolphthalein indicator turned into pink in the presence of bases, showing their basic nature. PROCEDURE 1 Place 1 mL of acetic acid, ★ Important Note hydrochloric acid, ammonium hydroxide, and sodium hydroxide in 4 separate test tubes and dilute ➔ Acids will turn blue litmus paper each with 1 mL of distilled water. into red, while bases will turn red Shake litmus paper into blue. ➔ Unionized methyl orange indicators OBSERVATIONS FOR PROCEDURE 1 will turn red in color in the presence of acids. Dip a clean glass rod into the solution and cautiously taste it. ➔ Phenolphthalein indicators will turn pink in basic solutions. Observation: Acetic acid and hydrochloric acid will have a sour taste, while ammonium hydroxide and sodium hydroxide will have a bitter taste. RELATIVE ACTIVITY OF ACIDS Test all the acids and bases above with red and blue litmus paper. APPARATUS USED Observation: Red litmus paper will turn ➔ Test tubes blue in the presence of bases (ammonium hydroxide and sodium hydroxide) and will remain red in the MATERIALS USED presence of acids (acetic acid and 1 mL of acetic acid hydrochloric acid); while blue litmus 1 mL of hydrochloric acid paper will turn red in the presence of 2 equal pieces of aluminum acids (acetic acid and hydrochloric acid) shaving and will remain blue in the presence of 2 equal pieces of marble chips bases (ammonium hydroxide and sodium hydroxide).  PROCEDURE 1 ➔ Hydrogen gas is evolved with To 1 mL of acetic acid and hydrochloric acid reacting faster hydrochloric acid in two separate with aluminum. test tubes add a piece of aluminum shaving (equal size) to each. ➔ Carbon dioxide is evolved with hydrochloric acid reacting faster What is the gas evolved? with marble chips. Answer: Hydrogen gas Which acid reacts faster with aluminum? Answer: Hydrochloric acid RELATIVE ACTIVITY OF BASES Write the equation involved. APPARATUS USED 2Al + 6HCl (Hydrochloric acid) -> 2AlCl3 ➔ Test tubes + 3H2 2Al + 6CH3COOH (Acetic Acid) -> MATERIALS USED 2Al(CH3COO)3 + 3H2 1 mL of ferric chloride 1 mL of aluminum sulfate 1 mL of zinc nitrate PROCEDURE 2 Dilute sodium hydroxide solution In separate test tubes, using the same acids above, place a piece of PROCEDURE 1 marble chips (same size) to each. Place 1 mL each of the ff. Solutions in 3 separate test tubes: ferric What is the gas evolved? chloride, aluminum sulfate and zinc Answer: Carbon dioxide nitrate. To each add slowly a few drops of dilute sodium hydroxide Which acid reacts faster with aluminum? solution until a precipitate is Answer: Hydrochloric acid obtained. Write the equation involved. OBSERVATIONS CaCO3 + 2HCl -> CaCl2 + H2O + CO2 Test tube 1: Ferric Chloride (FeCl3) + CaCO3 + 2CH3COOH -> Ca(CH3COO)2 + Sodium Hydroxide (NaOH) H2O + CO2 FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCL A reddish-brown precipitate is formed. ★ Important Note Test tube 2: Aluminum Sulfate (Al2(SO4)3) + Sodium Hydroxide (NaOH)  Al2(SO4)3 + 6NaOH -> 2Al(OH)3 + ➔ 2 Pipettes (5mL and 1mL) 3Na2SO4 ➔ Aspirator A white precipitate is formed. MATERIALS USED Test tube 3: Zinc Nitrate (Zn(NO3)2) + ➔ 0.1 M HCI Hydrogen Chloride (Acid) Sodium Hydroxide (NaOH) ➔ 0.1 M NaOH Sodium Hydroxide (Base) ➔ Methyl Orange and Alizarin Yellow Zn(NO3)2 + 2NaOH -> Zn(OH)2 + 2NaNO3 ➔ Distilled Water A white precipitate is formed. ➔ Litmus Paper EXPERIMENT 3 : HYDROGEN ION I. PREPARATION OF STANDARD ACID CONCENTRATION OF SOLUTIONS PROCEDURES A-D EXPERIMENT TOPIC OUTLINE ➔ Prepare four dry clean test tubes of I. The Preparation of Standard Acid the same size. Label them 1, 2, 3, & 4. II. Preparation of Standard Base Prepare your pipettes and aspirator. III. Comparison of Unknown with Ready your acid solution of HCl. Standard Acid/Base a - Pipette out 5mL of the 0.1 M HCl solution and place in test tube 1. TERMINOLOGY USED Label this as: [H] = 0.1 M ➔ Acid - “proton donor”; substance b - Pipette out 0.5mL of [H] = 0.1 M capable of donating a proton (H+ and place in test tube 2. Now, ion) to another substance; pH < 7; measure 4.5mL of distilled water releases H+ ions when dissolved in and add to test tube 2. Label this water as: [H] = 0.01 M ➔ Base - “proton acceptor”, a substance that accepts a proton c - Repeat process in (b) to test (H+ ion) from an acid; pH > 7; tube 3. Label it as: [H] = 0.001 M releases OH- ions when dissolved in d - Repeat process in (c) to test water tube 4. Label it as: [H] = 0.0001 M ➔ pH scale - stands for “potential of hydrogen”; measures how acidic or DILUTION FORMULA basic a substance is (0-14) ➔ molarity - concentration of a solution; number of moles of solute M1V1 = M2V2 per one liter of solution (mol/L) Where: M1 = initial molarity of stock solution V1 = initial volume of stock solution APPARATUS USED M2 = final molarity of new solution V2 = final volume of new solution ➔ 12 Test tubes  RECORDED COLORS OF HCl SOLUTIONS ➔ Methyl orange is a pH indicator that turns red when mixed with an acid, and turns yellow when with a base. ➔ Based on observations, here are the colors of each test tube: ◆ Test Tube 1 (a) - red ◆ Test Tube 2 (b) - light red ◆ Test Tube 2 (c) - orange SOLVING FOR FINAL CONCENTRATION ◆ Test Tube 2 (d) - yellow ➔ In procedures a-d, we are trying to solve for the final concentration or SOLVING FOR pH LEVEL molarity of the new diluted solution. ➔ To solve for pH, we use the formula: To find the M2 , we use the formula: pH = -log[H+] M2 = M1V1 / V2 a SOLUTION: a GIVEN: [H] = 0.1M HCl pH = -log[H+]a pH = -log[0.1M HCl] b GIVEN: SOLUTION: pH = 1 M1 = 0.1M M2 = M1V1 / V2 V1 = 0.5mL M2 = (0.1M)(0.5mL)/(5mL) b SOLUTION: M2 = ? M2 = 0.01M HCl pH = -log[H+]b V2 = 5mL pH = -log[0.01M HCl] pH = 2 c GIVEN: SOLUTION: M1 = 0.01M M2 = M1V1 / V2 c SOLUTION: V1 = 0.5mL M2 = (0.01M)(0.5mL)/(5mL) pH = -log[H+]c M2 = ? M2 = 0.001M HCl pH = -log[0.001M HCl] V2 = 5mL pH = 3 d GIVEN: SOLUTION: d SOLUTION: M1 = 0.001M M2 = M1V1 / V2 pH = -log[H+]d V1 = 0.5mL M2 = (0.001M)(0.5mL)/(5mL) pH = -log[0.0001M HCl] M2 = ? M2 = 0.0001M HCl pH = 4 V2 = 5mL CONCLUSION PROCEDURES E-G ➔ Add one drop of methyl orange (pH ➔ Acidic Solutions (like HCl): Diluting indicator) solution to each test tube. the solution increases the pH (i.e., ➔ Record the colors observed in each makes it less acidic). of the solutions for the various [H]. ➔ Calculate the pH of the standard solutions prepared in a, b, c, and d.  ➔ Calculate the pH of the standard II. PREPARATION OF STANDARD BASE solutions prepared in a, b, and c. PROCEDURES A-C RECORDED COLORS OF NaOH SOLUTIONS ➔ Prepare four dry clean test tubes of ➔ Alizarin yellow is a pH indicator that the same size. Label them 1, 2, 3, & 4. turns yellow in basic solutions and Prepare your pipettes and aspirator. red in acidic solution. Ready your base solution of NaOH. ➔ Based on observations, here are the colors of each test tube: a - [H] = 0.1 M ◆ Test Tube 1 (a) - light yellow ◆ Test Tube 2 (b) - even lighter b - [H] = 0.01 M yellow c - [H] = 0.001 M ◆ Test Tube 2 (c) - almost white d - [H] = 0.0001 M with a hint of yellow ◆ Test Tube 2 (d) - clear SOLVING FOR FINAL CONCENTRATION ➔ To find the M2 , we use the formula SOLVING FOR pOH LEVEL from the previous procedure. ➔ To solve for pOH, we use the formula: a GIVEN: [H] = 0.1M NaOH pOH = -log[OH-] b GIVEN: SOLUTION: M1 = 0.1M M2 = M1V1 / V2 V1 =0.5mL M2 = (0.1M)(0.5mL)/(5mL) a SOLUTION: M2 = ? M2 = 0.01M NaOH pOH = -log[OH-]a V2 = 5mL pOH = -log[0.1M NaOH] pOH = 1 c GIVEN: SOLUTION: M1 = 0.01M M2 = M1V1 / V2 b SOLUTION: V1 =0.5mL M2 = (0.01M)(0.5mL)/(5mL) pOH = -log[OH-]b M2 = ? M2 = 0.001M NaOH pOH = -log[0.01M NaOH] V2 = 5mL pOH = 2 d GIVEN: SOLUTION: c SOLUTION: M1 = 0.001M M2 = M1V1 / V2 pOH = -log[OH-]c V1 =0.5mL M2 = (0.001M)(0.5mL)/(5mL) pOH = -log[0.001M NaOH] M2 = ? M2 = 0.0001 NaOH pOH = 3 V2 = 5mL d SOLUTION: pOH = -log[OH-]d pOH = -log[0.0001M NaOH] PROCEDURES D-E pOH = 4 ➔ Add one drop of alizarin yellow (pH indicator) solution to each test tube. SOLVING FOR pH LEVEL ➔ Record the colors observed in each ➔ To solve for pH, we use the relation: of the solutions for the various [H].  pH + pOH = 14 ➔ Place a litmus paper in your solution to determine whether it is acidic or basic. a SOLUTION: ➔ Add one drop of methyl orange (if pH + pOH = 14 acidic) or alizarin yellow (if basic) to pH + 1 = 14 further determine its concentration. pH = 14 - 1 ➔ Record the H+ ion concentration of pH = 13 your unknown sample. b SOLUTION: pH + pOH = 14 pH + 2 = 14 DETERMINING THE UNKNOWN SAMPLE pH = 14 - 2 ➔ To determine what your unknown pH = 12 sample is, firstly, determine whether c SOLUTION: it is acidic or basic by using a litmus pH + pOH = 14 paper. Litmus paper turns red in pH + 3 = 14 acids and blue in bases. pH = 14 - 3 pH = 11 ➔ After finding whether it is an acid or base, find out its concentration by d SOLUTION: pH + pOH = 14 adding a drop of pH indicator. pH + 4 = 14 ➔ Recall that methyl orange goes from pH = 14 - 4 red to yellow, as the pH increases. pH = 10 Meanwhile, alizarin yellow goes from yellow to red, as pH increases. ➔ Compare color to the results of your CONCLUSION prepared standards and solve for pH using the formula: pH = -log[H⁺]. ➔ Basic Solutions (like NaOH): Diluting the solution decreases the pH (i.e., makes it less basic). SUMMARY ➔ Acidic Solutions: pH < 7 (e.g., 0.1 M HCl has a pH of 1.0). ➔ Neutral Solution: pH = 7 (e.g., pure III. COMPARISON OF UNKNOWN WITH water at 25°C). STANDARD ACID AND BASE ➔ Basic Solutions: pH > 7 (e.g., 0.1 M NaOH has a pH of 13.0). PROCEDURE ➔ Obtain 10mL of unknown sample from instructor. Pipette out 4.5mL of solution in a dry vial. PRACTICE PROBLEMS 1.What is the pH of a 0.01 M solution of hydrochloric acid (HCl)? Answer:  HCl is a strong acid and dissociates ○ [H⁺] = 10^(-pH) completely in water, so [H⁺] = 0.01 M. Given pH = 6: pH = -log[H⁺] ○ [H⁺] = 10^(-6) = 1.0 × 10^(-6) M pH = -log(0.01) = 2.0 2. If you dilute a 0.1 M solution of sodium hydroxide (NaOH) to 0.01 M, what will be the new pH of the solution? Answer: For NaOH, which is a strong base, [OH⁻] = 0.01 M after dilution. pOH = -log[OH⁻] = -log(0.01) = 2.0 pH = 14 - pOH = 14 - 2.0 = 12.0 3. How do you calculate the concentration of a hydrogen ion (H⁺) solution from a known pH value? Answer: To find the concentration of H⁺ from the pH, use the formula: [H⁺] = 10^(-pH) For example, if the pH is 3.5: [H⁺] = 10^(-3.5) ≈ 3.16 × 10^(-4) M 4. What is the pH of a solution if the concentration of hydroxide ions (OH⁻) is 0.001 M? Answer: First, find the pOH: ○ pOH = -log[OH⁻] = -log(0.001) = 3.0 Use the relationship pH + pOH = 14 to find the pH: ○ pH = 14 - pOH ○ pH = 14 - 3.0 = 11.0 5. A solution has a pH of 6. What is the concentration of hydrogen ions (H⁺) in this solution? Answer: To find the concentration of H⁺ from the pH, use the formula: