Chemistry Acid-Base Titration

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Chemistry: Strong Acid and Weak Base Titration Lab
Cherno Okafor
Mr. Huang
SCH4U7
November 21st, 2012

Data Collection and Processing

Concentration of the standard HCl solution: 0.1 M

Data Collection: | Trial 1| Trial 2| Trial 3|
Final HCl Buret Reading ± 0.05 mL | 38.3| 45| 54.5|
Initial HCl Buret Reading ± 0.05 mL | 29.9| 38.3| 45|
Volume of NaHCO3 used ± 0.1 mL | 9.2| 9.5| 9.8|

Qualitative Data:
* I used the methyl orange indicator which was suitable for my titration because of its clear and distinct colour change from orange to a bright red at the endpoint * At the beginning of the titration after I added 3 drops of methyl orange into the base (NaHCO3) and swirled, I began titrating the acid (HCl) slowly, and initially in the methyl orange and base, there was a tiny amount of red colour present, but then it quickly disappeared due to insufficient HCl (H+ ions)then I gradually kept titrating more acid while swirling and there was even more red colour present, until finally I reached the endpoint when the orange-yellow colour had completely transformed into a red colour * Changes from an orange-yellow colour (slightly higher pH 4.4) to a bright red colour (at low pH 3.1) at the endpoint and point of equivalence * Baking Soda (NaHCO3) absorbed the odour caused by the strong acid of HCl when I mixed the two: bleach-like smell

Processing
If the concentration of an acid or base is expressed in molarity, then the volume of the solution multiplied by its concentration is equal to the moles of the acid or base.

Therefore, the following relationship holds:

nVb x Cb = Va x Ca

Where: Vb = the volume of the base
Cb = the concentration of the base
Va = the volume of the acid
Ca = the concentration of the acid
n = the mole factor

In the case of hydrochloric acid and Sodium Bicarbonate (Baking Soda), the mole ratio is one to one, thus the mole factor is 1. Therefore, the volume of sodium bicarbonate multiplied by its concentration in molarity is equal to the moles sodium bicarbonate. The moles of sodium hydroxide are equal to the number of moles of hydrochloric acid in the reaction. The neutralization equation becomes: HCl + NaHCO3 NaCl + H2O + CO3 Hence, Cb = Va x Ca / Vb.

Trial 1 Calculation:
* First we need to find the change of volume of the acid used up in the titration: Va = Vfinal - Vintial
Va = 38.3 ± 0.05 – 29.9 ± 0.05
Va = 8.4 ± 0.1 mL

Therefore, nVb x Cb = Va x Ca
(1)(9.2 ± 0.1)(Cb) = (8.4 ± 0.1) (0.1 ± 0.0005) Cb = (8.4 ± 0.1) (0.1 ± 0.0005) / (9.2 ± 0.1) Cb = (8.4 ± 1.19%) (0.1 ± 0.5%) / (9.2 ± 1.09%) Cb = (0.84 ± 1.69%) / (9.2 ± 1.09%)

Cb = 0.0913 ± 2.78% 0.0913 ± 0.00254M is the concentration of the base for trial 1

Theoretical Base Concentration = 0.1 ± 0.0005 M
Experimental Base Concentration = 0.0913 ± 0.00254 M

Trial 2 Calculation:
* First find change of volume of the acid used up in the titration: Va = Vfinal – Vinitial
Va = 45 ± 0.05 – 36 ± 0.05
Va = 9.0 ± 0.1 mL

Therefore, nVb x Cb = Va x Ca
(1)(9.5 ± 0.1)(Cb) = (9.0 ± 0.1) (0.1 ± 0.0005) Cb = (9.0 ± 0.1) (0.1 ± 0.0005) / (9.5 ± 0.1) Cb = (9.0 ± 1.1%) (0.1 ± 0.5%) / (9.5 ± 1.05%) Cb = (0.9 ± 1.6%) / (9.5 ± 1.05%)

Cb = 0.0947 ± 2.65% 0.0947 ± 0.00251M is the concentration of the base for trial 2

Theoretical Base Concentration = 0.1 ± 0.0005 M
Experimental Base Concentration = 0.0947 ± 0.00251 M

Trial 3 Calculation:
* First find change of volume of the acid used up in the titration: Va = Vfinal – Vinitial
Va = 54.5 ± 0.05 – 45 ± 0.05
Va = 9.5 ± 0.1 mL

Therefore, nVb x Cb = Va x Ca
(1)(9.8 ± 0.1)(Cb) = (9.5 ± 0.1)...
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