Identify Unknown Solutions PDF

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Identify Unknown Solutions

There are three unknown solutions in test tubes at your station. It is your goal for this part to identify the solutions, collecting data to support your conclusions. Notes : You will not receive more solution, so make sure to use only a portion at a time if you would like to run multiple tests. In addition to your unknowns, there is an NaOH solution provided (in a jar by the TA). You may only use things that are listed on the equipment list or in common areas of the lab. Litmus Purple - then neutral Red- acidic Blue - basic

Note - strong acid 0-1 in pH Weak acid 1-7 pH Take it out and do a little mini titration 1. On your answer sheet, create a data section with the tests , etc that you perform. ● Have to be able to determine the difference between seawater, tap water, and deionized water: neutral pH ● Do conductivity tests of water - the seawater will have the most ionic content, then tap water, then DI water ● sea water - 1000ish ● Tap water - 100ish to 300 ● DI water - very little i. Use conductivity probe 2. In a short paragraph for each unknown, identify it and support your conclusion with your data. Procedure: 1. Test the pH of solutions using pH paper a. If the pH is neutral (around 7, 6-8): i. Can be water or EDTA ii. Do conductivity test (make sure units in ppm) - calibrate probe 1. ~1000 - seawater 2. ~200 to 500 - tap water or EDTA 3. Very little - DI or EDTA iii. If very little

1. 2. 3. 4.

Put some eriochrome black T (3 drops) and a little bit of NaOH If you observe a color change, then it’s EDTA. (blue) If no color change (just purple) - it is DI water Do this test with a small portion of the sample of water and a small portion of the NaOH! b. if the pH is higher than 8-9 it is basic, either 0.1 and 1.0 M I. do the conductivity test, testing both the NaOH given and the unknown sample II. If the unknown sample and NaOH have the same (roughly) conductivity, then they are the same concentration - same amounts of ions in solution III. 1.0 M would have higher conductivity than 0.1 M so use the NaOH given as a reference IV. if concentration of NaOH = 0.5, multiply conductivity by 2 to get conductivity of 1.0 M solution - if the conductivity of the unknown is much less, than the unknown would be the 0.1 M concentration strong base C. If the pH is low, below 6 then it will be an acid - a titration with NaOH will be performed pH of strong acid around 0, 0.1 of a strong acid around 1, 2-3 weak acid - trouble distinguishing? This is where you do a titration I. Very low pH - one or below means this will be a strong acid, if above then probably weak acid 1. If given two acidic solutions, one with greater conductivity is the higher concentration 2. Can perform a titration here using NaOH 3. Strong acid, strong base equivalence point will be pH=7 4. Can use bromothymol blue indicator 5. M 1V 1, = M 2 V 2 solving for M 2 = molarity of the acid 6. if the indicator changes color with the same amount of base added (ex 2 mL base neutralizes 2 mL of acid) then they are the same concentration (for strong acid, strong base) - just do this!! Add same amount of base to chosen amount of acid, if it neutralizes, they’re the same concentration, if not then not ii. Middle acidic pH (3-5) - this is either a 1.0 M or 0.1 M weak acid 1. If conductivity is similar to that of NaOH, it’s a strong acid 2. If conductivity is less than NaOH - it’s a weak acid 3. Using a graduated pipette, remove some of the unknown and place it in a beaker - dilute with water to cover the bottom of the beaker - how does this affect the calculations though? - i think just use volume of acid put in, not including DI water used to dilute 4. Add NaOH to the burette Add phenolphthalein indicator 5. 6. Titrate solution and calculate the molarity of the acid using M 1V 1, = M 2 V

solving for M 2 = molarity of the acid What’s the V acid if we diluted it here though? - just use v of acid not including DI water!!!! 2

7.

Further Notes: ● Use flask for dilution ● Relative conductivity of EDTA - WILL UPDATE - apparently a little less than tap water? ● Need only one trial ● Record observations From titration: - mL of NaOH titrated - moles of NaOH titrated - assume 1:1 reaction with acid - = moles of acid titrated - [H+] = 10^-pH - Ka = [A-][H+]/[HA] @ equivalence, pH = pKa Don’t rely on just one test - do a couple of other tests

Write down all volumes Make sure you write the observations - change in temperature, conductivity Show how you calculated qrxn - density of water is 1 g/mL then, you can calculate mass

Tests pH paper (acidic, neutral, or basic) Conductivity test (ppm) Eriochrome Black T (indicator)

Unknown 1

Unknown 2

Unknown 3

Titration with NaOH: mL acid added Mol NaOH (=mole acid) mL NaOH used acid] = mole NaOH/L acid

Conductivity probe needs to be in ppm

Experiment → change units → ppm 1 M naoh - do conductivity - if you have a base with around the same conductivity...