A. Given Kw for water is 2.4x10-14 at 37 deg C. What is the pH of a natural aque
ID: 914794 • Letter: A
Question
A. Given Kw for water is 2.4x10-14 at 37 deg C. What is the pH of a natural aqueous solution at 37 deg C?B. If Ka of a weak acid is 5.0x10-6. What is the pH of a 0.43 M solution of this acid?
C. What concentration of SO3 2- is in equilibrium with Ag2SO3(s) and 7.60x10-3 M Ag+?
A. Given Kw for water is 2.4x10-14 at 37 deg C. What is the pH of a natural aqueous solution at 37 deg C?
B. If Ka of a weak acid is 5.0x10-6. What is the pH of a 0.43 M solution of this acid?
C. What concentration of SO3 2- is in equilibrium with Ag2SO3(s) and 7.60x10-3 M Ag+?
B. If Ka of a weak acid is 5.0x10-6. What is the pH of a 0.43 M solution of this acid?
C. What concentration of SO3 2- is in equilibrium with Ag2SO3(s) and 7.60x10-3 M Ag+?
Explanation / Answer
Ans A:
Expression for Kw for water can be written as follows:
Kw = [H+][OH-]
In natural water [H+]= [OH-]
Therefore, Kw = [H+][H+] = [H+]2
Substitute Kw = 2.4 x 10-14
2.4 x 10-14 = [H+]2
[H+] = (2.4 x 10-14)1/2
[H+] = 1.549 x 10-7
Expression for pH can be written as follows:
pH = -log[H+]
pH = -log (1.549 x 10-7 )
pH = 6.810
Therefore, pH of natural aqueous solution is 6.180
Ans B:
Expression for Ka of weak acid HA can be written as follows:
Ka = [H+][A-] / [HA]
Since, [H+] = [A-]
Ka = [H+][H+] / [HA]
Ka = [H+]2 / [HA]
Substitute Ka = 5.0x10-6 and [HA] = 0.43 M
5.0 x 10-6 = [H+]2 / 0.43
[H+]2 = 5.0 x 10-6 x 0.43
[H+]2 =2.15 x 10-6
[H+] =1.47 x 10-3
Expression for pH can be written as follows:
pH = -log[H+]
pH = -log (1.47 x 10-3 )
pH = 2.83
Therefore pH of a 0.43 M solution of given acid is 2.83
Ans C:
Expression for solubility product of Ag2SO3 can be written as follows:
Ksp = [Ag+]2 [SO3]
Ksp for Ag2SO3 is 1.50 x 10-14
Let s be the moles of Ag2SO3 dissociated to give 2s moles of Ag+ and s moles of SO32-then, Ksp can be written as follows:
Ksp = (2s)2 x s
Since, Ag+ already present is 7.60 x 10-3 M
Therefore, total [Ag+] is 2s + 7.60 x 10-3 M
Hence, Ksp can be written as follows:
Ksp = (2s + 7.60x10-3)2 x s
Substitute value of Ksp = 1.50 x 10-14
1.50 x 10-14 = (2s + 7.60x10-3)2 x s
Assuming s to be very very small, 2s can be neglected.
1.50 x 10-14 = (7.60x10-3)2 x s
s = (1.50 x 10-14) / (7.60x10-3)2
s = (1.50 x 10-14) / (5.776 x 10-5)
s = 0.260 x 10-9 M
Therefore, 0.260 x 10-9 M SO32- is in equilibrium with Ag2SO3(s)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.