Sample Problem. A 0.1355 g sample (uncorrected mass) of dry, primary-standard Na2C03 required 25.05 mL of HCI solution (at 23.3 °C) to reach the end point. A blank titration with the same amount of indicator required 0.04 mL. Use linear interpolation and the data in Table 2-7 of Harris to estimate the density of water at 23.3 °C. What is the concentration of the HCI solution, corrected to 20 °C? [Buoyancy correction factor: mtrue/mread = 1.000 3244] [Interpolated density of water at 23.3 °C: 0.997 468 9 g/mL] [HCI concentration at 20 °C: 0.102343 M] M buoyancy Corr. factor = m me mread = (1- 0.0012 g/mL 8.0 gimi 0.0012 9/mc 2.63 gimu 1.0003244 1 0.99985 0.9995257 density interpolation of water low 23°C 0.9975415 g/m² + meas. 23.3.C -0.9974689 g/mL) high-24.c 0.9972995. gliml dint = 0.99754154 + (23.3 - 23) · 69972995 - 24 23 0.9975415

Sample Problem. A 0.1355 g sample (uncorrected mass) of dry, primary-standard Na2C03 required 25.05 mL of HCI solution (at 23.3 °C) to reach the end point. A blank titration with the same amount of indicator required 0.04 mL. Use linear interpolation and the data in Table 2-7 of Harris to estimate the density of water at 23.3 °C. What is the concentration of the HCI solution, corrected to 20 °C? [Buoyancy correction factor: mtrue/mread = 1.000 3244] [Interpolated density of water at 23.3 °C: 0.997 468 9 g/mL] [HCI concentration at 20 °C: 0.102343 M] M buoyancy Corr. factor = m me mread = (1- 0.0012 g/mL 8.0 gimi 0.0012 9/mc 2.63 gimu 1.0003244 1 0.99985 0.9995257 density interpolation of water low 23°C 0.9975415 g/m² + meas. 23.3.C -0.9974689 g/mL) high-24.c 0.9972995. gliml dint = 0.99754154 + (23.3 - 23) · 69972995 - 24 23 0.9975415

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter22: Bulk Electrolysis: Electrogravimetry And Coulometry

Section: Chapter Questions

Problem 22.33QAP

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