Titration Of Sulfuric Acid With Sodium Hydroxide 3y3f2n

Titration of Sulfuric Acid with Sodium Hydroxide By: Jayden Blackwood Date: 6/1/2014 Ms. Psarianos – Chemistry Grade 11 Question: What is the concentration of H2SO4 Sample #86, based on how much NaOH (concentration of 0.100 mol dm-3) is needed to neutralize it?

Data Collection and Processing:

Given Concentration of Sodium hydroxide (NaOH): The given concentration of Sodium hydroxide in this titration is 0.100 mol dm-3 Raw Data Tables:

Qualitative Raw Data 

Both the Sodium hydroxide (NaOH) and Sulfuric acid (H 2SO4) have a clear, translucent color.  When Bromothymol blue was added to the un-titrated Sulfuric acid, it became light yellow in color. When the titration was complete, the solution of combined acid and base was a light green color. A higher concentration of base resulted in the solution taking on a light blue color.

Data Recorded from Titration Trial # 1 2 3 4 5

Amount of Sulfuric acid (H2SO4) ( mL ± 0.06) 25 25 25 25 25

Amount of Sodium hydroxide (NaOH) delivered (mL ± 0.05) 14.70 14.20 14.50 14.30 14.35

Processing of Raw Data (Calculations and Presentation of Result): The balanced chemical equation of this titration is as follows. 2NaOH + H2SO4  (Na)2SO4 + 2H2O Ionic Equation Na+ + OH- + 2H+ + (SO4)2-  Na+ + (SO4)2- + 2H2O Net Ionic Equation  H+ + OH-  H2O

Main Calculations: Mean volume of NaOH delivered:

14.7 mL ±0.05+14.2 mL ± 0.05+14.5 mL ± 0.05+14.3 mL ± 0.05+14.35 mL ± 0.05 5

¿

¿

72.05 ± 0.25 5

The absolute uncertainties were added together, and then converted to a percentage.

72.05 ± 0.35 54

¿ 14.4 mL ± 0.35

Therefore the mean volume of NaOH delivered is 14.4 mL ± 0.35%. Calculations for the unknown concentration of H2SO4:

The balanced chemical equation of this titration stated earlier is: 2NaOH + H2SO4  (Na)2SO4 + 2H2O Finding the number of moles of NaOH: −3

0.01441 L ± 0.35 ×0.1000 mol d m =number of moles of NaOH

To isolate for the number of moles, the molarity formula (c = n/v) was re arranged.

¿ 0.00144 mol ± 0.35

The mole ratio will now be applied with regards to the H2SO4,

0.001441mol ± 0.35 2

There as twice as much of moles of NaOH than H2SO4, therefore the number of moles of NaOH is divided by two.

¿ 0.000721 mol ±0.35

Solving for Molar concentration of H2SO4,

0.0007205mol ± 0.35 0.025 L ± 0.24

The percentage uncertainty of the volume of H2SO4 was calculated by solving

0.06 25 . The measurement in

−3

¿ 0.0288 mol d m ± 0.59

Therefore the unknown concentration of H2SO4 is 0.0288 mol dm-3 ± 0.59% (percent) or 0.0002 (absolute).

Conclusion and Evaluation:

In conclusion, the unknown concentration of H2SO4 is 0.0288 mol dm-3±0.0002. The pre determined and calculated theoretical value of the concentration of H 2SO4 given the initial measurements is 0.029 mol dm-3. The aforementioned result obtained through calculation lies within the range of the calculated uncertainty (0.029 to 0.0286). The calculated percentage error of concentration is as follows:

0.029−0.0288 0.029 ¿ 0.69 In reality this is a very low percentage error, giving the obtained answer more validity. Both systematic and random errors were observed.

Limitations/Effects/Suggested Improvements:

Limitation

Effect on result

Suggested Improvements

Sodium hydroxide solutions are somewhat unstable. They tend to absorb atmospheric carbon dioxide. (systematic error)

There could have potentially been uned carbonates in the final solution if the Sulfuric acid did not expel all of the atmospheric CO2.

The indicator used (bromothymol blue) is slightly acidic in nature (systematic error)

This would increase the acidity of the final solution and could have potentially changed the results of the titration.

Sulfuric acid will slowly expel carbon dioxide from the solution, but the initial presence of carbonates means that to reach the final titrated solution we need to add excess of the titrant (NaOH). Bromothymol blue should be applied to the point where the color of the solution is apparent. Acidity of the indicator can be minimized. Perhaps another indicator could have been used like phenolphthalein which would have given the solution a pink color in substitute of bromothymol blue.

When the lab was conducted the right shade of green is unknown (random error)

It was difficult to pour the NaOH due to the faulty spout. Lots of NaOH spilt, as supposed to having a consistent flow (random error)

Because the correct color of the completely titrated solution is not known it is difficult to judge whether or not the shade obtained is the correct color of green. Because of this error perhaps too much NaOH may have split into the final solution, resulting in the slight percentage error of 0.69%

A photo or shade of paper outlining the fully titrated solution using bromothymol blue would aid in determining whether or not the solution is fully titrated. The Sodium hydroxide could have been placed in a better container, or the spout could be cleaned for any debris that may have blocked the opening