Please check if my second data table is correct. I don't think the molar mass is correct Part II. Determination Ka and Molar Mass of Unknown Monoprotic AcidAssignment # 21Mass of the acid, in stock solution, g4.452Volume of the stock solution, ml250.003Molarity of the standardized NaOH solution, M0.09Data Table 1. Titration Data1Volume of the Unknown acid used for titration, mL10.00Volume of NaOH at equivalence point from first derivative, ml3 Volume of NaOH at equivalence point from second derivative, mL222.1322.134Average volume of NaOH for calculations22.13Initial pH3.086pH at equivalence point from Titration Curve Page8.766pH at half-equivalence point from Titration Curve Page4.937pH at half-equivalence point from the Graph below9.15Data Table 2. Final Results of the Titration of Unknown AcidMoles of NaOH at equivalence point, mol|Moles of Unknown acid at equivalence point, molMolarity of Unknown acid, MMoles of Unknown acid in the stock solution, molMolar Mass of Unknown acid, g/molpH at half-equivalence point2.02E-032.02E-038.09E-032.02E-0374.08344.937pKa4.938Ка1.17E-05According to Appendix II Unknown acid is identified asPropanoic Acid2. Titration Curve - pH vs Volume13.0012.0011.0010.009.008.007.006.005.004.003.002.001.000.000.005.0010.0015.0020.0025.0030.0035.00Volume of NaOH, mlEquation 1 (Eq 1)y= 0.08x+4.0474Slope (ml)0.0800Intercept (b1)4.0474Equation 2 (Eq 2)y= 4.804x-97.314Slope (m2)4.804Intercept (b2)-97.31Equation 3 (Eq 3)y=0.062x-10.348Slope (m2)0.0620Intercept (b3)10.348V1, Volume of NAOH, ml at intercept of Eq 1 and Eq 221.46(b2 - b1) (m1 - m2)v2, Volume of NAOH, ml at intercept of Eq 2 and Eq 322.70(b3 - b1) (m2 - m3)Veg, Volume of NAOH, mL at equivalence point(V2 • V1) /222.08pHeq, pH at equivalence point8.76m2 x Veq • b2Vhalf-eq, Volume of NaOH, ml at half - equivalence point11.04Veq 12pH half-eq, pH at half - equivalence point4.93m1x Vhalf-eq • b1pKa4.93Ка1.17E-05Hd

The unknown acid is monoprotic, also the base used (NaOH) is monoacidic.

Science

Chemistry

Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acio

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter16: Reactions Between Acids And Bases

Section: Chapter Questions

Problem 16.54QE

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Please check if my second data table is correct. I don't think the molar mass is correct

Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acid
Assignment # 2
1
Mass of the acid, in stock solution, g
4.45
2
Volume of the stock solution, ml
250.00
3
Molarity of the standardized NaOH solution, M
0.09
Data Table 1. Titration Data
1
Volume of the Unknown acid used for titration, mL
10.00
Volume of NaOH at equivalence point from first derivative, ml
3 Volume of NaOH at equivalence point from second derivative, mL
2
22.13
22.13
4
Average volume of NaOH for calculations
22.13
Initial pH
3.08
6
pH at equivalence point from Titration Curve Page
8.76
6
pH at half-equivalence point from Titration Curve Page
4.93
7
pH at half-equivalence point from the Graph below
9.15
Data Table 2. Final Results of the Titration of Unknown Acid
Moles of NaOH at equivalence point, mol|
Moles of Unknown acid at equivalence point, mol
Molarity of Unknown acid, M
Moles of Unknown acid in the stock solution, mol
Molar Mass of Unknown acid, g/mol
pH at half-equivalence point
2.02E-03
2.02E-03
8.09E-03
2.02E-03
74.08
3
4
4.93
7
pKa
4.93
8
Ка
1.17E-05
According to Appendix II Unknown acid is identified as
Propanoic Acid
2.

Titration Curve - pH vs Volume
13.00
12.00
11.00
10.00
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
Volume of NaOH, ml
Equation 1 (Eq 1)
y= 0.08x+4.0474
Slope (ml)
0.0800
Intercept (b1)
4.0474
Equation 2 (Eq 2)
y= 4.804x-97.314
Slope (m2)
4.804
Intercept (b2)
-97.31
Equation 3 (Eq 3)
y=0.062x-10.348
Slope (m2)
0.0620
Intercept (b3)
10.348
V1, Volume of NAOH, ml at intercept of Eq 1 and Eq 2
21.46
(b2 - b1) (m1 - m2)
v2, Volume of NAOH, ml at intercept of Eq 2 and Eq 3
22.70
(b3 - b1) (m2 - m3)
Veg, Volume of NAOH, mL at equivalence point
(V2 • V1) /2
22.08
pHeq, pH at equivalence point
8.76
m2 x Veq • b2
Vhalf-eq, Volume of NaOH, ml at half - equivalence point
11.04
Veq 12
pH half-eq, pH at half - equivalence point
4.93
m1x Vhalf-eq • b1
pKa
4.93
Ка
1.17E-05
Hd

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