Sample+lab+writeup--Chemistry+x


 * Title:** Titration Lab


 * Purpose:** To measure the concentration ( in moles/ liter) of a HCl solution by titration against a .0711 mole/ liter NaOH solution


 * Procedure:**
 * 1) Rinse a buret with about 5 ml of the NaOH (aq), discard the solution.
 * 2) Fill the buret almost to the top of the gradations with the NaOH (aq).
 * 3) Measure and record the initial volume of the buret (±.01ml).
 * 4) Deliver 10.00 (±.04) ml HCl (aq) to a 100 ml beaker with a 10.00 ml volumetric pipette and a rubber bulb
 * 5) Add 5 or 6 drops phenolphthalein indicator
 * 6) Add NaOH (aq) from the buret to the beaker, slowing down towards the end, until the final drop allows the pink color to persist after swirling.
 * 7) Measure and record the final volume of the buret (±.01ml).
 * 8) Repeat steps 2-7 until three trials agree in volume of NaOH (aq) required to titrate within ±.2 ml

(calculated) || 39.10 || 39.13 || 39.31 || 38.84 ||
 * Data** (copy) Data taken Period 3, 1/7/09
 * Trial || 1 || 2 || 3 || 4 ||
 * Initial volume (ml) || 0.81 || 1.48 || 0.28 || 0.92 ||
 * Final volume (ml) || 39.91 || 40.61 || 39.59 || 39.76 ||
 * Volume(ml) NaOH used
 * Volume(ml) NaOH used


 * Calculations:**

Volume (ml) NaOH used =final volume-initial volume**=**39.91 ml-0.81 ml
 * =39.10 ml (Trial 1)**
 * =39.13 ml (Trial 2)**
 * =39.31 ml (Trial 3)**
 * =38.83 ml (Trial 4)**

Average volume NaOH (aq) = (39.10 ml +39.13 ml +39.31 ml +38.84 ml )/4 = **39.10 ml**

Volume NaOH (aq) (L)= 39.10 ml x (1x10-3L / 1 ml) = **.03910 L**

Moles NaOH used= .03910 L x (.0711 mole NaOH / 1 L)= **.00278 mol NaOH**

Balanced Reaction: **NaOH (aq) + HCl (aq) →H 2 O (l) + NaCl (aq)**

Moles HCl neutralized= .00278 mol NaOH x 1HCl/1NaOH=**.00278 mol HCl**

Volume HCl (aq) (L)= 10.00 ml x (1x10-3L / 1 ml) = **0.01000 L**

Concentration of HCl =.00278 mol HCl / 0.01000 L = **.278 mol / L**

Analysis: The four trials used varied in volume ±.3 ml from the average, or about .8%. This indicates an uncertainty in the conclusion of about the same .8%, or ±.002 mol/L. Consistent error could have been introduced by contamination, though no indications were observed.

**
 * Conclusion:** The HCl solution contained **.278 ±.002 moles HCl / L