not sure what I am doing here can someone please help. Part I1 -Sodium Acetate 0
ID: 1028128 • Letter: N
Question
not sure what I am doing here can someone please help.
Part I1 -Sodium Acetate 0.3 M 0.5M 52 3.31 X Io 35 X Equilibrium Expression for Sodium Acetate: For the 0.3 M Sodium Acetate Solution: I(nitial) C(hange E(equilibrium) O. 3 Kb Calculation 3.65 YIo For the 0.5 M Sodium Acetate Solution: I(nitial) C(hange) E(equilibrium) 0.5 13 Kb Calculation Average Kb of both solutions -13 K for water (@L22XDc) What is the percent error of your Kw versus the actual Kw for water at the temperature you measured? (Estimate the actual Kw value for your temperature from the table of Kw values) 27Explanation / Answer
Part II
Equilibrium expression for sodium acetate
CH3COO- + H2O <==> CH3COOH + OH-
Kb = [CH3COO-][OH-]/[CH3COO-]
--
pH of 0.3 M CH3COONa solution
amount of sodium acetate hydrolyzed in water = [OH-] = 3.31 x 10^-8 M
CH3COO- + H2O <=====> CH3COOH + OH-
I 0.3 - - -
C -3.31 x 10^-8 - +3.31 x 10^-8 +3.31 x 10^-8
E almost 0.3 - 3.31 x 10^-8 3.31 x 10^-8
So,
Kb = (3.31 x 10^-8)^2/0.3
= 3.65 x 10^-15
--
pH of 0.5 M CH3COONa solution
amount of sodium acetate hydrolyzed in water = [OH-] = 3.5 x 10^-7 M
CH3COO- + H2O <=====> CH3COOH + OH-
I 0.5 - - -
C -3.5 x 10^-7 - +3.5 x 10^-7 +3.5 x 10^-7
E almost 0.5 - 3.5 x 10^-7 3.5 x 10^-7
So,
Kb = (3.5 x 10^-7)^2/0.5
= 2.45 x 10^-13
--
average Kb = 1.24 x 10^-13
Kb for water given = 1.22 x 10^-19
So,
percent error = (1.24 x 10^-13 - 1.22 x 10^-19) x 100/1.24 x 10^-13 = 99.99%
the value of Kb is well off of the actual value.
The error is coming from the incorrect measurement of pH of the solution in Table, Part II at the top of page.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.