# Condumetric and Gravemetric Lab Report

Topics: Mole, Concentration, Solubility Pages: 3 (602 words) Published: September 5, 2011
Conductimetric Titration and Gravimetric Determination of a Precipitate Objective:
* Measure the conductivity of the reaction between sulfuric acid and barium hydroxide * use conductivity values to determine equivalence point * measure mass of a product to determine equivalence point gravimetrically * calculate molar concentration of barium hydroxide solution Procedure:

* First, combine 10.0 mL of the Ba(OH)2 solution with 50 mL of distilled water. Then, measure out 60 mL of 0.100 M H2SO4. Set up a conductivity probe and open programs by connecting to logger pro. After that, start to titrate with increments of 1.0 mL. Keep titrating with smaller increments until it is pretty close to the 100 microsiemens/cm mark. Calculate the lowest value. After that, record the volume as the equivalence point. Finally, filter and measure the mass of the barium sulfate precipitate. Data:

| Trial 1| Trial 2| | | |
Equivalence Point (mL)| 8.0| 7.6| | | |
Mass of filter paper and precipitate (g)| | 1.2417| | | | Mass of filter paper (g)| | 1.0469| | | |
Mass of precipitate (g)| | 0.1948| | | |
Molarity of H2SO4 (M)| 0.100| 0.100| | | |

Calculations and graphs:
* see last page for graph
* Equations:
* M = n/V
* Molar Mass = m/n
* percent error = [|(experimental – actual)|/actual] • 100 1.
7.6 mL| 1L| 0.100mol|
| 1000 mL| 1L|
= 7.6 • 10-4 mol

2. mol H2SO4 = mol Ba(OH)2

7.6 • 10-4 mol Ba(OH)2|
0.010 L |
= 0.076 M

3.

0.1948 g BaSO4| 1 mol|
| 233.43 g|
= 0.0008351 mol BaSO4

4. mol BaSO4 = mol Ba(OH)2

0.0008351 mol Ba(OH)2|
0.010 L|
= 0.0835 M

5.
(|0.076 – 0.100|)|
0.100|
* 100 = 24%

(|0.0835 – 0.100|)|
0.100|
* 100 = 17%

Discussion of Theory:
* The concept performed in this lab was determining molar concentration of conductimetric titration and gravimetric determination of a...