16) 30.0 mL of 0.1 M Na-CO, solution was titrated against HCI (10 mL) to reach the bromocresol end point. What is the concentration of HCI solution in mol/L? a) 0.60 b) 0.30 17) 250 ml ENCI c) 0.03 d) 0.60
16) 30.0 mL of 0.1 M Na-CO, solution was titrated against HCI (10 mL) to reach the bromocresol end point. What is the concentration of HCI solution in mol/L? a) 0.60 b) 0.30 17) 250 ml ENCI c) 0.03 d) 0.60
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
Chapter12: Solutions
Section: Chapter Questions
Problem 12.107QE
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Q 16 please
![11) The concentration of the solution of sodium thiosulfate that was standardized using
0.2210 g of potassium iodate KIO (MW-214.00 g/mol) in water with excess KI and H₂SO,
for acidification. The liberated I was titrated with 41.64 ml. of sodium thiosulfate.
31₂ +3K; SO+ 3H₂O
KIO+SKI + 3 H₂SO
S.O,¹ +21
25:0 +1
a) 0.149 M
b) 0.123 M
c) 0.235 M
d) 0.325 M
12) If 15.00 ml. of 0.1M NaOH solution is needed to reach the end point of 10 mL. H₂PO4.
Then the concentration of H₂PO, is: [M.wt. of H.PO.- 98 g/mol]
a) 6.533
b) 14.699 g/l.
c) 7.349 g/L
d) 18.762 g
The following DATA for determination of calcium and magnesium in water by titration with
EDTA. USE THIS DATAT FOR 13 AND 14
The volume of (Ca²+Mg¹2) solution
The volume of EDTA solution( using EBT indicator)
The volume of EDTA solution (using murexide indicator)
The molarity of EDTA
[M.wt. of Ca 40.0/Mg-24.3 g/mol]
13) The mass in gram of Ca in the 10 ml solution is :
a) 1.6
b) 0.85
14). The mass of Mg" expressed in ppm is:
b) 16
a) 8.5
c) 0.085
c) 0.16
10.00 mL
15.00 mL
8.00 mL
0.5 M
d)16
d) 1.86 ml
d) 0.16
15) The volume of 0.015 M HCl solution should be used to prepare 100 ml of 5.23 x10³ M HCl
solution is.
a) 2.87 ml
b) 0.286 ml
c) 3.40 ml
16) 30.0 mL of 0.1 M Na,CO, solution was titrated against HCI (10 mL) to reach the bromocresol end
point. What is the concentration of HCI solution in mol/L?
a) 0.60
b) 0.30
c) 0.03
d) 0.60
17) 25.0 mL of NaCl unknown concentration was titrated against 23.0 mL of 0.009 M AgNO3.
Calculate the ppm of NaCl in the original sample. (Na-23 g/mol, Cl- 35.5 g/mol)
a) 484.38
b) 293.94
c) 48.44
d) 29.39](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa82c98f-d3ff-46ff-9a06-9e0a8e58c6be%2F28e71dd4-5199-4982-b30f-e58201452e86%2F9epn7r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:11) The concentration of the solution of sodium thiosulfate that was standardized using
0.2210 g of potassium iodate KIO (MW-214.00 g/mol) in water with excess KI and H₂SO,
for acidification. The liberated I was titrated with 41.64 ml. of sodium thiosulfate.
31₂ +3K; SO+ 3H₂O
KIO+SKI + 3 H₂SO
S.O,¹ +21
25:0 +1
a) 0.149 M
b) 0.123 M
c) 0.235 M
d) 0.325 M
12) If 15.00 ml. of 0.1M NaOH solution is needed to reach the end point of 10 mL. H₂PO4.
Then the concentration of H₂PO, is: [M.wt. of H.PO.- 98 g/mol]
a) 6.533
b) 14.699 g/l.
c) 7.349 g/L
d) 18.762 g
The following DATA for determination of calcium and magnesium in water by titration with
EDTA. USE THIS DATAT FOR 13 AND 14
The volume of (Ca²+Mg¹2) solution
The volume of EDTA solution( using EBT indicator)
The volume of EDTA solution (using murexide indicator)
The molarity of EDTA
[M.wt. of Ca 40.0/Mg-24.3 g/mol]
13) The mass in gram of Ca in the 10 ml solution is :
a) 1.6
b) 0.85
14). The mass of Mg" expressed in ppm is:
b) 16
a) 8.5
c) 0.085
c) 0.16
10.00 mL
15.00 mL
8.00 mL
0.5 M
d)16
d) 1.86 ml
d) 0.16
15) The volume of 0.015 M HCl solution should be used to prepare 100 ml of 5.23 x10³ M HCl
solution is.
a) 2.87 ml
b) 0.286 ml
c) 3.40 ml
16) 30.0 mL of 0.1 M Na,CO, solution was titrated against HCI (10 mL) to reach the bromocresol end
point. What is the concentration of HCI solution in mol/L?
a) 0.60
b) 0.30
c) 0.03
d) 0.60
17) 25.0 mL of NaCl unknown concentration was titrated against 23.0 mL of 0.009 M AgNO3.
Calculate the ppm of NaCl in the original sample. (Na-23 g/mol, Cl- 35.5 g/mol)
a) 484.38
b) 293.94
c) 48.44
d) 29.39
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