Discuss three factors that influence the distribution of chemical species in an
ID: 490397 • Letter: D
Question
Discuss three factors that influence the distribution of chemical species in an aquatic system. In each ease give an example or illustration to support your discussion Acetic acid (CH_3COOH) has a dissociation constant of 1.8 times 10^-5. What would be the composition of a 0.10M acetic acid in a solution of pH 4.7? Below is a distribution diagram for phosphoric acid in an aqueous medium, showing the fractions of the various phosphate species as a function of pH. What would be the fractional values of the phosphate species in a lake of pH 4.3? If the total phosphate content of the lake is 62 mu g/L, what would be the concentrations of the phosphate species at pH 4.3? The distribution of phosphate in aqueous media is governed by the equilibria shown below. Estimate from the diagram the values of K_1, K_2, and K_3, indicating what these k values mean. H_3PO_4 doubleheadarrow H^+ + H_2PO_4^-, K_1 = ? H_2PO_4 doubleheadarrow H^+ + HPO_4^2-, k_2 = ? HPO_4^2- doubleheadarrow H^+ + PO_4^3-, K_3 = ?Explanation / Answer
Answer(B)
Given:
Molarity of acetic acid = 0.1 M
Dissociation constant = Ka = 1.8x10-5
Solution:
CH3COOH + H2O ? CH3COO- +H3O+
Ka = [H3O+][CH3COO-]/[CH3COOH]
1.8x10-5 = x2 / (0.1- x)
x is very negligible compare to initial concentration of acid.
Hence,
1.8x10-5 = x2 / 0.1
x2 = 1.8 x 10-5 x 0.1
x =[H3O+] = 0.001342 M acetic acid
As [H+] = 0.001342 M
pH = -log10[H+]
pH= - log10(0.001342)
pH= 2.87
But the pH of solution is 4.7 so
X = epH = 0.0090 so composition of acetic acid wil be
0.1 -X = 0.1 - 0.0090 = 0.091 M
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.