07-13: The diprotic acid Hal. has Kor 1.4 x 10% and Ka 1.2 x 10-¹0 97-Calculate the molarity of H* of 0.080 M HL solution 13-107 B)4.110 C) 1.3 * 10% D) 4.1 * 107 08-Calculate the molarity of Lof 0.080 M H₂L solution A) 1.4-10 B) 1.2 10 C) 14x10 D) 1.2 × 10-10 09-Find the pH of a solution prepared by dissolving 0.85 g of NaHL (188.1 g/mol) and 1.05 g of NasL (194.0 g/mol) in 250.0 mL of water. A) 10.84 B) 9.00 C) 7.84 D) 10.0 Q10- The fraction of L2 (a of L2) at pH-12.00 is A) 0.923 B) 0.813 C) 0.992 D) 0.089 Q11- At pH-5.85 A) [HL]=(H₂L] D) [L²] = [HL] Q12- At pH=7.89 A) [H₂L]=[L²] D) [L²] = [HL] (1661 g/mol) and 1.05 g of B) [L]> [H₂L] B) [HL]=[H₂L] C) [H₂L]>[HL] C) [H₂L]> [HL] SILT
07-13: The diprotic acid Hal. has Kor 1.4 x 10% and Ka 1.2 x 10-¹0 97-Calculate the molarity of H* of 0.080 M HL solution 13-107 B)4.110 C) 1.3 * 10% D) 4.1 * 107 08-Calculate the molarity of Lof 0.080 M H₂L solution A) 1.4-10 B) 1.2 10 C) 14x10 D) 1.2 × 10-10 09-Find the pH of a solution prepared by dissolving 0.85 g of NaHL (188.1 g/mol) and 1.05 g of NasL (194.0 g/mol) in 250.0 mL of water. A) 10.84 B) 9.00 C) 7.84 D) 10.0 Q10- The fraction of L2 (a of L2) at pH-12.00 is A) 0.923 B) 0.813 C) 0.992 D) 0.089 Q11- At pH-5.85 A) [HL]=(H₂L] D) [L²] = [HL] Q12- At pH=7.89 A) [H₂L]=[L²] D) [L²] = [HL] (1661 g/mol) and 1.05 g of B) [L]> [H₂L] B) [HL]=[H₂L] C) [H₂L]>[HL] C) [H₂L]> [HL] SILT
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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pleasssseee solve question 9
![07-13: The diprotic acid Hil. has Kor 1.4 x 10% and Ka 1.2 x 10-10
97-Calculate the molarity of H* of 0.080 M HL solution
13-107 B)4.1-10 C) 1.3×10% D) 4.1 * 107
08-Calculate the molarity of Lof 0.080 M H₂L solution
A) 1410 B) 1.2 10 C) 14x10 D) 1.2 × 10-10
09-Find the pH of a solution prepared by dissolving 0.85 g of NaHL (188.1 g/mol) and 1.05 g
of NazL (194.0 g/mol) in 250.0 mL of water.
A) 10.84
B) 9.00
C) 7.84
D) 10.0
Q10- The fraction of L2 (a of L2) at pH-12.00 is
A) 0.923 B) 0.813 C) 0.992 D) 0.089
Q11- At pH-5.85
A) [HL]=[H₂L]
B) [L]> [H₂L]
D) [L²] = [HL]
Q12- At pH=7.89
A) [H₂L]=[L²
B) [HL]=[H₂L]
D) [L²] = [HL]
CILL (1661 g/mol) and 1.05 g of
C) [H₂L]>[HL]
C) [H₂L]> [HL]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbccdf819-703a-40ae-b479-32ba98a4eafe%2Fa19db827-051c-4b7e-b5b7-b08684021ff8%2Fzr23uc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:07-13: The diprotic acid Hil. has Kor 1.4 x 10% and Ka 1.2 x 10-10
97-Calculate the molarity of H* of 0.080 M HL solution
13-107 B)4.1-10 C) 1.3×10% D) 4.1 * 107
08-Calculate the molarity of Lof 0.080 M H₂L solution
A) 1410 B) 1.2 10 C) 14x10 D) 1.2 × 10-10
09-Find the pH of a solution prepared by dissolving 0.85 g of NaHL (188.1 g/mol) and 1.05 g
of NazL (194.0 g/mol) in 250.0 mL of water.
A) 10.84
B) 9.00
C) 7.84
D) 10.0
Q10- The fraction of L2 (a of L2) at pH-12.00 is
A) 0.923 B) 0.813 C) 0.992 D) 0.089
Q11- At pH-5.85
A) [HL]=[H₂L]
B) [L]> [H₂L]
D) [L²] = [HL]
Q12- At pH=7.89
A) [H₂L]=[L²
B) [HL]=[H₂L]
D) [L²] = [HL]
CILL (1661 g/mol) and 1.05 g of
C) [H₂L]>[HL]
C) [H₂L]> [HL]
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