lab5report - + ion, so this eliminates H 2 SO 4 . In order...

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We began the first rough titration with the 0.0979M NaOH with the unknown concentration of unknown acid K. The initial pH was 2.35 and the final pH was 11.5 after 13mL of NaOH. The first big leap in pH occurred between 5mL and 6mL, resulting in pH 3.03 and pH 9.85, respectively. We estimated the amount needed to reach the equivalence point to be around 5.5mL. We then began the more accurate titration by stopping at 3mL of NaOH and titrating by 0.1mL increments. At about 5.8mL titrated was the enormous leap in pH that was predicted to be at 5.5mL. We continued to titrate by 0.1mL intervals until we reached 7mL titrated and continued with 0.5mL increments. After 13mL of NaOH, the final pH was 11.53. We can first see that acid K is a strong acid because from the titration curve the pH is at 7 and a strong acid and strong base produce neutral salts. Therefore, it cannot be any weak acid. It is also has only one equivalence point, indicating that it forms only one H
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Unformatted text preview: + ion, so this eliminates H 2 SO 4 . In order to find the concentration of the acid, we must use the known molarity of the NaOH solution and the volume of HCl used in the following equations: = M1V1 M2V2 . = . ( . ) M10 05L HCl 0 0979M NaOH 0 0059L = . M1 0 01M HCl From the chart, one can see that the equivalence point is at pH 7. Since NaOH is a strong base, this indicates that K must be a strong acid. Therefore, the only acid choices remain are either hydrochloric acid or sulfuric acid. Because sulfuric acid is diprotic and our acid was not, the only choice remaining is HCl. In comparison with other groups in the lab, we found that we had the same acid as the group that had acid C, except in a lower concentration. It took the other group 10mL to reach the equivalence point when it only took us approximately 5.9mL NaOH. Their molarity was approximately 0.02M HCl, while ours was 0.01M HCl....
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lab5report - + ion, so this eliminates H 2 SO 4 . In order...

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