DATA-
Standardization of NaOH Solution
Standardization of HCL Solution
Concentration of Unknown HCL Solution……….Unknown 2
AVERAGE MOLARITY .041 M
Conclusion- The titration methods were use to prepare solutions of acid and base and also to determine the unknown concentration of an acid solution. Average Molarity of the unknown solution was .041 M.
Supplementary Questions-
1. Using the exact mass of NaOH measured and recorded at the beginning of the experiment, calculate the molarity 250 mL NaOH solution. How does this compare to the NaOH concentration you calculated using the data in part 4? What could be some reasons for this?
1g * 1mol/40g * 1L/.25 L= .1 M It is greater then the concentration in part 4.
2. Why were the amounts of distilled water added to the acids not precisely recorded and considered to be irrelevant? This might be because distilled water has no effect on the amount of the acid because the acid is stronger.
3. Would the calculated molarity of the HCL solutions be higher or lower or not affected if each of the following occurred. Explain your answers
A. The buret containing NaOH was rinsed with distilled water but not rinsed with NaOH before being filled.
Reduce NaOH concentration, more NaOH would be used. Therefore molarity of HCL would be more.
B. The calculated molarity was too high
The moles of NaOH used would be high. Therefore HCL would be high
C. The tip of the NaOH buret contained an air bubble at the beginning of the titration but not at the end.
It would seem like more NaOH was used Therefore HCL higher.
4. Suppose you used the base Ca(OH)2 instead of NaOH. What changes would need be made to the calculations of the HCL concentrations?
It would have to double because twice as more OH ions are used.
