Question: Molarity HCl = .353 mol/L Trial 1 Sample 1 2 3 4 5 Temp at Decantation 60 50 40.2 30 21.2 Vol. HCl used to
Molarity HCl = .353 mol/L
| Trial 1 | |||||
| Sample | 1 | 2 | 3 | 4 | 5 |
| Temp at Decantation | 60 | 50 | 40.2 | 30 | 21.2 |
| Vol. HCl used to reach end | 16.50 | 15.8 | 14.5 | 7.4 | 2.88 |
| Trial 2 | |||||
| Temp at Decantation | 61 | 50 | 38.0 | n/s | 22.0 |
| Vol HCl used to reach end | 16.50 | 12.99 | 14.3 | n/s | 2.66 |
| Avg Temp | 60.50 | 50 | 39.10 | 30.00 | 21.60 |
| Avg Vol | 16.50 | 14.40 | 14.4 | 7.4 | 2.77 |
| Molarity Tetraborate anion |
Calculate the molarity of B4O5(OH)4^2- (tetraborate anion) for each column using the average volme of HCl at the equivalence point and the average molarity of HCl which is : .353 mol/L
Remember that 5.0 mL of the tetraborate solution was titrated.
B4O5(OH)4^2-(aq) + 2 H+(aq) + 3 H2O(l) --> 4 H3BO3(aq)
Also.. how do you calculate Ksp?
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Answer 1 Molarity of B4O5OH42 tetraborate anion Trial 1 Sample12345 Avg Vol HCl1650144014474277 Mola... View full answer
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