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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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