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Titration Expt. Overview Day 1: You will create a ~0.1 M NaOH solution Day 2: You will standardize your solution of NaOH by titrating it with a solid acid,

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Presentation on theme: "Titration Expt. Overview Day 1: You will create a ~0.1 M NaOH solution Day 2: You will standardize your solution of NaOH by titrating it with a solid acid,"— Presentation transcript:

1 Titration Expt. Overview Day 1: You will create a ~0.1 M NaOH solution Day 2: You will standardize your solution of NaOH by titrating it with a solid acid, KHP Purpose: To verify your [NaOH] up to 3 SF It should be close to M NaOH Day 3: You will titrate your now known concentration of NaOH with an unknown concentration of HCl Purpose: To determine the [HCl]

2 Day 1: Making a NaOH solution You and your partner will make 250 mL of a ~0.1 M NaOH solution How will you do this? What do you need to know? What equipment do you need? Remember, M = mol/L Calculate the mol NaOH necessary  g NaOH Measure g NaOH using a balance and dissolve in mL distilled water together in a volumetric flask Calculate and record actual [NaOH] in M (mol/L) Should be close to M, but won’t be exact 3 SF

3 Day 2: Standardization of NaOH (aq) Purpose: To verify your [NaOH] by reacting (titrating) it with an known acid, KHP Buret w/ base ~0.1 M NaOH Flask w/ acid and PHTH ind. ~0.3 g KHP in ~75 mL H 2 O Faint pink tinge… Use white paper underneath to see better (PHTH)

4 Day 2: Standardization of NaOH (aq) Purpose: To verify your [NaOH] by reacting (titrating) it with an known acid, KHP To neutralize… mol H + = mol OH - Acid = ~0.3 g solid KHP (potassium hydrogen phthlate) mass g KHC 8 H 4 O 4 (to 3 SF)  mol KHP = mol H + Recall M = mol/L, so mol OH - = M OH · L OH therefore… mol H + = M OH · L OH Do one dry run  ball park amount NaOH to add Do at least two good trials, if not three Average your [NaOH] results

5 Day 3: Titration of HCl (aq) with NaOH (aq) Purpose: To determine unknown [HCl] by reacting (titrating) it with known base, NaOH Buret w/ base ~0.1 M NaOH Flask w/ acid and ind. Yellow ~ 15 mL of ? M HCl Green Endpoint Use white paper underneath to see better Blue Overshoot (too basic) Bromothymol blue

6 Day 3: Titration of HCl (aq) with NaOH (aq) Purpose: To determine unknown [HCl] by reacting (titrating) it with known base, NaOH To neutralize… mol H + = mol OH - Recall M = mol/L … M H+ · L H+ = M OH- · L OH- Do one dry run  ball park amount NaOH Do at least three good trials Whatever you don’t finish today, you can finish tomorrow Average your [HCl] results

7 Lab Report Typed (calculations may be hand-written) Sections Overview (3-5 sentences) Procedure Day 1: Making NaOH solution Day 2: Titration (Standardization) with KHP Day 3: Titration with HCl Data Tables Show organized data for all three days of expt. (include dry run data and label as such) Include related data table with each day’s procedure Calculations Show all calculations (no need to use dry run data) Include related calculations with each day’s procedure Written portions should be: 3 rd person Passive voice Past tense

8 Lab Report Organization (/50 pts) Overview (3-5 sentences) Day 1: Making NaOH solution Procedure Data table Calcs Day 2: Titration (Standardization) with KHP Procedure Data table Calcs Day 3: Titration with HCl Procedure Data table Calcs

9 Suggested Point Breakdown(/50 pts) Overview (3-5 sentences) (+5?) Procedures (+15, +5 each) Data Tables: Days 2 and 3 (+10, +5 each) Calculations: Days 2 and 3 (+20, +10 each) Grade calculations/data first for all labs, and then go back through and grade written portions.

10 strong acid + strong base  salt + HOH product pH = 7… use bromothymol blue yellow (acid), green (neutral), blue (base) HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l) [H + ] = 0.50 M [OH - ] = ? M ~ mL V B Measure both V in expt. (to the nearest 0.00 mL) Trials 1.(Dry Run)… calculations unimportant (ball park numbers) Calculate [OH - ] for trials 2-4 and then average

11 Strong Acid/Strong Base Titration pH Equivalence Pt mol H + = mol OH - mL of base HCl + NaOH  H 2 O + NaCl At this point, everything is water, Na + and Cl -, so the pH = 7 Endpoint Overshoot Titrate with Bromthymol Blue

12 Weak Acid/Strong Base Titration

13 Weak acid + strong base  weak base + HOH product pH > 7… use phenolphthalein (PHTH) clear  acid pink  base HCH 3 COO (aq) + NaOH (aq)  NaCH 3 COO (aq) + HOH (l) (acetic acid, 0.24 M (Na + (aq) + i.e. vinegar) V B CH 3 COO - (aq)) [H + ] = ? ~ 3-5 mL Trials 1.(Dry Run)… calculations unimportant (ball park numbers)

14 Equivalence Pt mol H + = mol OH - mL of base At this point, the solution contains water, Na +, and C 2 H 3 O 2 -, so the pH > 7. Endpoint Overshoot BUFFERING REGION HC 2 H 3 O 2 C 2 H 3 O H + Weak Acid/Strong Base Titration HC 2 H 3 O 2 + NaOH  H 2 O + NaC 2 H 3 O 2 Titrate with Phenolphthalein pH

15 Equivalence Pt mol H + = mol OH - mL of acid At this point, the solution contains water, Cl -, and NH 4 +, so the pH < 7. Endpoint Overshoot NH 3 + H + NH 4 + BUFFERING REGION Weak Base/Strong Acid Titration HCl + NH 4 OH  H 2 O + NH 4 Cl Titrate with Methyl Orange pH


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