How the Dilution of a Strong Acid Affects the Overall pH of the Solution

Topics: Acid, Sodium hydroxide, Hydrochloric acid Pages: 2 (545 words) Published: August 2, 2013
Reactions in aqueous solutions

Purpose: The purpose of the lab was to see how the dilution of a strong acid affected the overall PH of the solution. This was also the case with the strong base.

Data:
 | PH| H+| OH-| Experimental PH | Percent error|
HCl 1| 1.7| 0.0125| 8E-13| 1.90| 10.67|
HCL 2| 2.87| 6.25E-04| 1.6E-11| 3.20| 10.43|
HCl 3| 3.2| 4.69E-05| 2.1322E-10| 4.33| 26.08|
HCl 4| 3.4| 4.69E-06| 2.1322E-09| 5.33| 36.20|
NaOH 1| 12.35| 8E-13| 0.0125| 12.10| 2.09|
NaOH 2| 11.15| 1.6E-11| 6.25E-04| 10.80| 3.28|
NaOH 3| 10.25| 2.1322E-10| 4.69E-05| 9.67| 5.99|
NaOH 4| 9.35| 2.1322E-09| 4.69E-06| 8.67| 7.83|
Com 1| 2.2| 0.0000001| 0.0000001| 7.00| 68.57|
Com 2| 3.49| 0.0000001| 0.0000001| 7.00| 50.14|
Com 3| 3.85| 0.0000001| 0.0000001| 7.00| 45.00|
Com 4| 4.01| 0.0000001| 0.0000001| 7.00| 42.71|
D.I.| 5.9| 0.0000001| 0.0000001| 7.00| 15.71|
Figure: 1

Balanced equation: HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)
Experimental Yield: .423g of NaCl
Theoretical Yield: .242g of NaCl
Percent Yield: 175%

Calculations:
(.5M)(1ml)/(25ml)=0.0125 M of HCl
-log(0.0125)=1.90 PH
14-1.90=12.1 POH
(0.5M)(.01L)(1mol of NaCL/1mol of HCl)(58.44g/1mol)=0.242g of NaCl produced

Conclusion
The purpose of this experiment was to see how the dilution of a strong acid and base affected the PH of those solutions. From the data in table 1, as more water is used to dilute the HCL solution the PH starts to rise. It can be inferred that as more water is added the PH will start to plateau at 7, which is the PH of pure water. The same will happen with the NaOH solution. As more water is added the PH starts to approach the PH of 7 and it can be inferred that as more water is added it will come to reach the PH of 7. In regards to the combined results it was a little surprising that the PH results were not closer to that of a neutral PH of 7. Theoretically if equal...
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