1) Answer the following question based on a 0.002 M solution of chloroacetic aci

Question

1) Answer the following question based on a 0.002 M solution of chloroacetic acid.
Ka = 1.4x10-3

a) Calculate the equilibrium concentration of CH2ClCOOH, H3O+, CH2ClCOO-

b) Calculate the percent ionization of this acid.

c) Calculate pH of the solution.

Calculate the Kb value for the chloroacetate ion (CH2ClCOO-).

e) Calculate the equilibrium concentration for all species present (except water).

f) What is the pH of this solution

a) I got [H3O+] = [CH2ClCOO-] = 1.67x10-3

    [CH2ClCOOH]: 0.002-( 1.67x10-3 ) = 3.3x10-4

b) (1.67x10-3 / 0.002) x 100 = 83.5% (Too High according to the 5% rule that my Prof. told my class)

c) [H3O+] = [CH2ClCOO-] = 1.11x10-3

    [CH2ClCOOH]: 0.002-( 1.11x10-3 ) = 8.9x10-4

pH= -log[H3O+] = -log( 1.11x10-3 ) = 2.95

d)This is where i got stuck, Hopefully my answer above were right also.

Thank you in advance for helping me.

Explanation / Answer

I'll explain the idea then I'll correct what you got wrong

a) Ka = [H3O+][CH3ClCOO-]/[CH3ClCOOH]

If Ka = 1.4e-3 then

Ka = 1.4e-3 = (X)(X)/(0.002-x)

Will solve the long way to compare your answer with mine

2.8e-6 - 0.0014 - x^2 = 0

Quadratic formula yields root of 1.114e-3

I got [H3O+] = [CH2ClCOO-] = 1.67x10-3 (Too high compared to what I got, the general rule of thumb is; if the concentration you start with (i.e. 0.002M) is greater than 100*Ka then you can neglect x on the bottom as an approximation. You got your answer by doing this, but you can't in this case because it isn't significantly higher than Ka)

[CH2ClCOOH]: 0.002-( 1.67x10-3 ) = 3.3x10-4 (Fixed below)

0.002 - 1.114e-3 = 0.0008862

b) (1.67x10-3 / 0.002) x 100 = 83.5% (Too High according to the 5% rule that my Prof. told my class)

With mine, 8.86e-4/0.002 = 44.31% (still high for a weak acid like you said, but much more reasonable, I would think that the values your professor gave for Ka are the reason why it's so high, Ka should have been smaller)

Though an acid with a Ka like that has a pKa of 2.854 which is almost as strong as H3O+ to be honest, so this isn't a weak acid by any means. Consider your answer reasonable math wise.

c) [H3O+] = [CH2ClCOO-] = 1.11x10-3
[CH2ClCOOH]: 0.002-( 1.11x10-3 ) = 8.9x10-4
pH= -log[H3O+] = -log( 1.11x10-3 ) = 2.95 (Can't use this approximation since its a "weak acid" better to use henderson hasselbach since it partially dissociates)

pH = pKa + log[A]/[HA]

pH = -log(1.4e-3) + log([1.114e-3]/[0.002]) = 2.6

d) pKa + pKb = 14

pKa = 2.854

14-2.854 = pKb = 11.146

e) Now this is going to get crazy, if we still use 0.002M for CH3ClCOOH with such a small value of Kb our values are going to be nuts (I'll show you what work I can but I would double check before turning in). Sorry I can't be more specific but the directions are a little confusing

Solve backwards using Kb

Kb = 10^-pKb

Kb = 10^-11.146 = 7.145e-12

7.145e-12 = [0.002][0.002]/(X-0.002)

[OH-] ,[CH3ClCOOH] = 0.002 M

X = [CH3ClCOO-] = 559832 M

f) pOH = pKb + log[OH-]/[BOH]

pOH = 11.146 + log(0.002)/(559832)

pOH = 11.145

pH + pOH = 14

pH = 2.855

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