Expl-HCIStandardization-5.pd Tools Fill & Sign Comment Open 1 50% Sample Problem

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Expl-HCIStandardization-5.pd Tools Fill & Sign Comment Open 1 50% Sample Problems. The molar mass and density of Na2CO3 are given above. For this first experiment, please show your calculation of the buoyancy correction factor with the sample problems. Always calculate buoyancy correction factors to at least 5 decimal places. You will not have to show this calculation in future experiments, although you will always have to write its numerical value and show how you used it to find the corrected mass. The same holds for linear interpolation of water densities: we ask you to show the full calculation this time, but in future experiments you can just write its numerical value and show how you used it to correct for thermal expansion. Make sure your answers exactly match the ones given below. We will often include extra guard digits to help you check this A 0.1355 g sample (uncorrected mass) of dry, primary-standard Na2CO3 required 25.05 mL of HCl solution (at 23.3 oC) to reach the end point. A blank titration with the same amount of indicator required 0.04 m L. Use near interpolation and the data in Table 2-7 of Harris to estimate the density of water at 23.3 oC. What is the concentration of the HCl solution, corrected to 20 OC? [Buoyancy correction factor: m read 1.000 3244 Interpolated density of water at 23.3 oC: 0.997 468 9 g/mLJ IHCI concentration at 20 OC: 0.102343 M] Questions for Further Thought. You do not have to answer these questions in your report, but some of them would be good exam questions once we get to chapters 10 and 11 of U 8.50 x 11.00 in

Explanation / Answer

Answer: Here we know that at the end point the number of moles of acid is equal to the number of moles of base

and hence the number of moles of sodium carbonates = mass / molar mass = 0.1355 / 105.9

= 0.001279mol

And hence the number of moles of acid is also equal to 0.001279 mol at end point

Now the requiered concentration is = 0.001279 * 1000 / 25.05 = 0.051M

and hece by using all the defects and we get the required concentration = 0.05M

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