A solution of citric acid (H3CgH;O,) with a known concentration of 0.200 M H;C,H;O, is titrated with a 0.750 M N2OH solution. How many ml of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H;CGH;O, , according to the following balanced chemical equation: H;CgH;O, + 3 NaOH – Na;CgH;0, + 3 H20
A solution of citric acid (H3CgH;O,) with a known concentration of 0.200 M H;C,H;O, is titrated with a 0.750 M N2OH solution. How many ml of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H;CGH;O, , according to the following balanced chemical equation: H;CgH;O, + 3 NaOH – Na;CgH;0, + 3 H20
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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A solution of citric acid (H3C6H507) with a known concentration of 0.200
M H;C,H;0, is titrated with a 0.750 M NaOH solution. How many ml of NAOH
are required to reach the third equivalence point with a starting volume of 25.0
mL H;C6H;0, , according to the following balanced chemical equation:
H3C6H5O, + 3
NaOH – Na;C,H;O, +
3 H20
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
11.25
0.200
0.020
2.00 x 10*
2.22
2
0.001
3
6.67
1
1000
0.750
25.0
20.0
15.0
L NAOH
mol H3CgH;O7
LH;CGH;O, mL H;C6H;O,
g NaOH
mL NaOH
M H;C;H;07
M NaOH
g H;C6H;O,
mol NaOH
9,523
3
étv](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87adfcfd-efb1-42d9-9318-c7f4e6e0bd19%2F372b15dd-73e7-4d51-810c-7f7a017fc9f2%2F4ap6v0p_processed.png&w=3840&q=75)
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* ? 94%
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b Answered: What is the activati
101 Chem101
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Update :
Question 6 of 47
Submit
A solution of citric acid (H3C6H507) with a known concentration of 0.200
M H;C,H;0, is titrated with a 0.750 M NaOH solution. How many ml of NAOH
are required to reach the third equivalence point with a starting volume of 25.0
mL H;C6H;0, , according to the following balanced chemical equation:
H3C6H5O, + 3
NaOH – Na;C,H;O, +
3 H20
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
11.25
0.200
0.020
2.00 x 10*
2.22
2
0.001
3
6.67
1
1000
0.750
25.0
20.0
15.0
L NAOH
mol H3CgH;O7
LH;CGH;O, mL H;C6H;O,
g NaOH
mL NaOH
M H;C;H;07
M NaOH
g H;C6H;O,
mol NaOH
9,523
3
étv
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