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Answer to Question #8622 in Organic Chemistry for Imani

Question #8622
a solution is created by dissolving 1.935g of sodium acetate, NaCH3CO2,in 850.0mL of .405M acetic acid,CH3COOH,at 25'C.Ka for acetic acid is 1.8*10^-5 at 25'C. assume that the volume of the solution does not change. find the concertratioof ion,CH3COO- in the final solution
Expert's answer
CH3COOH <-> H+ + CH3COO-

Ka = [H+][CH3COO-]/[CH3COOH]

CH3COONa <-> CH3COO- + Na+

Sodium acetate is a strong electrolyte and breaks up into ions completely

1.935 n
CH3COONa <-> CH3COO- + Na+
82

n = 0.0236 mol
C = 0.0236/0.85 = 0.0278 M (mol/L)

using that Ka of final solution is:

Ka = [H+][0.0278 + x]/0.405 = 1.8*10-5
[H+] = x too because CH3COOH <-> H+ + CH3COO-

x*(0.0278+x) = 1.8*10-5 * 0.405
x^2 + 0.0278x - 7,29*10^-6 = 0
x = 2.59*10-4 M
final C of CH3COO- = 0.0278 + 0.000259 = 0.02806 M

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