Help! Ka=1.8*10^-5 These are equations 10&11 concentration of HC2H302, M measure

Question

Help!
Ka=1.8*10^-5 These are equations 10&11 concentration of HC2H302, M measured pH theoretical pH 1.0 x 10 1.0 x 10 1.0 x 10 1.0 x 10 calculated Ka of HC2H3O2 literature K, of HC2H3O2 based on pH data concentration of HC2H3O2, M 1.0 x 10 1.0 x 10 1.0 x 10 1.0 x 10 Ill. Preparing HC,H30, 2. Calculate the theoretical pH of each HC2H302 solution, using Equation 5 Solutions and and the Ka value given in Table 1. 3. K. for using your measured pH values for each solution and the model of Equations 10 and 11. Assume that the Determining pH concentration of undissociated HC2H3O2 is the same as its initial concentration. 4. Consult Table 1 to find the reported Ka for HC2H3O2.

Explanation / Answer

2. Calculate pH

HC2H3O2 = 1 x 10^-1 M

Ka = [C2H3O2-][H3O+]/[HC2H3O2]

1.8 x 10^-5 = x^2/1 x 10^-1

x = [H3O+] = 1.34 x 10^-3 M

pH = -log[H3O+] = 2.87

HC2H3O2 = 1 x 10^-2 M

Ka = [C2H3O2-][H3O+]/[HC2H3O2]

1.8 x 10^-5 = x^2/1 x 10^-2

x = [H3O+] = 4.24 x 10^-4 M

pH = -log[H3O+] = 3.37

HC2H3O2 = 1 x 10^-3 M

Ka = [C2H3O2-][H3O+]/[HC2H3O2]

1.8 x 10^-5 = x^2/1 x 10^-3

x = [H3O+] = 1.34 x 10^-4 M

pH = -log[H3O+] = 3.87

HC2H3O2 = 1 x 10^-4M

Ka = [C2H3O2-][H3O+]/[HC2H3O2]

1.8 x 10^-5 = x^2/1 x 10^-4

x = [H3O+] = 4.24 x 10^-5 M

pH = -log[H3O+] = 4.37

3. Ka values from calculated pH,

HC2H3O2 = 1 x 10^-1 M

Ka = (1.34 x 10^-3)^2/(1 x 10^-1 - 1.34 x 10^-3)

     = 1.82 x 10^-5

HC2H3O2 = 1 x 10^-2 M

Ka = (4.24 x 10^-4)^2/(1 x 10^-2 - 4.24 x 10^-4)

     = 1.88 x 10^-5

HC2H3O2 = 1 x 10^-3 M

Ka = (1.34 x 10^-4)^2/(1 x 10^-3 - 1.34 x 10^-4)

     = 2.07 x 10^-5

HC2H3O2 = 1 x 10^-4 M

Ka = (4.24 x 10^-5)^2/(1 x 10^-4 - 4.24 x 10^-5)

     = 3.12 x 10-5

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