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
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
What is the concentration of the HCl solution, corrected to 20 deg C? The correct answer is given: 0.102343 M, but I cannot get that exact answer
![Sample Problem.
A 0.1355 g sample (uncorrected mass) of dry, primary-standard Na2CO3 required
25.05 mL of HCl 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]
= (-
buoyancy
Corr. factor
M me
mread
Jon
meas.
23.3 (
density interpolation of water
low -23°c
high-24.c
1.0003244
JM
0.0012 g/mL
9.0 gimi
0.0012 g/m²
1
-
JOY
0.99985
0.9995257
dint= 0.99754154 + 23.3- 23
24-23
€
0.9975415 g/m²+
- (0.9974689 g/mL
0.9972995
glme
.
e
1.9972995-
0.9975415
dint = 0.997 541 54 + (0.3) • (-2.42 x 10-+)
dint
0.9974689 gim
=
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8afcb4f4-6b38-4217-bf1b-cd0eebc89fe1%2F0975021c-ae80-4e92-a499-18ed4961e80e%2Fa213fvs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Sample Problem.
A 0.1355 g sample (uncorrected mass) of dry, primary-standard Na2CO3 required
25.05 mL of HCl 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]
= (-
buoyancy
Corr. factor
M me
mread
Jon
meas.
23.3 (
density interpolation of water
low -23°c
high-24.c
1.0003244
JM
0.0012 g/mL
9.0 gimi
0.0012 g/m²
1
-
JOY
0.99985
0.9995257
dint= 0.99754154 + 23.3- 23
24-23
€
0.9975415 g/m²+
- (0.9974689 g/mL
0.9972995
glme
.
e
1.9972995-
0.9975415
dint = 0.997 541 54 + (0.3) • (-2.42 x 10-+)
dint
0.9974689 gim
=
-
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