Neutralization Lab

Analysis:
1. Sample Calculation of Condition 1 Trial 1:
Total Volume – Initial Volume = Neutralized Volume
9.5 – 5.0 = 4.5. Therefore 4.5 mL of NaOH was needed to neutralize HCl.

Table 3: The amount of NaOH (mL) needed to neutralize HCl in Condition 1, 2 , and 3 and their 3 trials.

Trial 1
(mL)
Trial 2
(mL)
Trial 3
(mL)
Condition 1
4.5
4.0
4.0
Condition 2
8.0
8.0
9.0
Condition 3
9.5
9.0
9.5

2. Sample Average Calculation of Condition 1:

Table 4: The average volume (mL) of HCl & NaOH in each Condition

Average Volume of HCl & NaOH
Condition 1
9.2
Condition 2
16.5
Condition 3
19.3
Figure 1: Bar Graph comparing the average amount of NaOH and HCl needed for Neutralization in each Condition

4. HCl + NaOH  H2O + NaOH Already Balanced

Conclusion:
In Conclusion, the effect of increasing the volume of HCl, is that the amount of NaOH needed to neutralize HCl is the almost the initial volume of HCl. In Condition 2 Trial 2 the initial volume of HCl was 8.0 mL, the substance stayed pink at 16.0 mL, so the amount of NaOH needed for neutralization was 8.0 mL. In the other conditions and trials the most that base was off from the acid was 1 mL. The average HCl and NaOH stayed fairly close, this is shown in Fig 1. The Hypothesis was supported.

Evaluation:
Limitation/Design Error
Specific Effect(s) on Data
Way to fix it?
The phenolphthalein dropper was not completely accurate
Spit out different amount of solution each time. It effected the mass, increased it.
Pipettes could be used, they won’t have different amount of solution come out each time
Stirring Rod
Some of the solution got stuck to the tip of the rod because it’s hollow. Decreased the mass
Use a Stirring Rod that isn’t hallow so that no solution gets left behind on it.