The compound ethanolamine, HOCH2CH2NH2, is a weak base. If 49.5 mL of a 0.029 M

ID: 831215 • Letter: T

Question

The compound ethanolamine, HOCH2CH2NH2, is a weak base. If 49.5 mL of a 0.029 M solution of ethanolamine is titrated with 0.100 M hydrochloric acid..

3.) What would be the PH at the beginning of this titration? (Kb=3.1*10^-5)

4.) How many mL of HCL will be needed to reach the equivalence point?

5.) What would be the pH at the equivalence point?

6.) What would be the pH when 10 more mL of hydrochloric acid was added past the equivalence point?

Please show all work (as much as possible). I am really struggling with understanding titration problems. Thanks so much for all your help! :-)

3) HOCH2CH2NH2 + H2O <------> HOCH2CH2NH3+ + OH-

initial moles = 0.029 x 49.5/1000 = 0.0014355 , 0 0

at equi 0.0014355-x x x

Kb = [HOCH2CH2NH3+][OH-]/[HOCH2CH2NH2]

3.1 x 10^-5 = ( x/0.0495)^2/( 0.0014355-x)/0.0495

x = 0.0000462 = OH- moles , [OH-] = (0.0000462/0.0495) = 0.0009394

pOH = -log ( 0.0009394) =4 , pH = 14-4 = 10

4) at equivalence point HCl moles = base moles

0.1 x V = 0.0014355 , V = 0.014355 L = 14.36 ml

5) HOCH2CH2NH2+ H+ <----> HOCH2CH2NH3+

at equi we have HOCH2CH2NH2 all consumed and hence moles of HOCH2CH2NH3+ = 0.014355

now back reaction occurs , total vol = 0.0495+0.014355 = 0.064

HOCH2CH2NH3+ <----> HOCH2CH2NH2 + H +

Ka = [HOCH2CH2NH2][H+]/[HOCH2CH2NH3+]

Ka = Kw/Kb(base) = 10^-14/3.1x10^-5 = 3.226 x 10^-10

Ka = 3.226 x 10^-10 = ( x/0.064)^2 /( 0.0014355-x)/0.064

x = 1.722x 10^-7 = H+ moles , [H+] = ( 1.722 x 10^-7 /0.064) = 2.69 x 10^-6

pH = -log ( 2.69 x 10^-6) = 5.57

6) when 10 ml more is added past equivalence point we have excess acid

total vol = 24.36+49.5 = 74 ml = 0.074 L

excess H+ = ( 0.024355 x 0.1 - 0.0014355) = 0.001

[H+] = ( 0.001/0.074) = 0.0946

pH = -log ( 0.0946) = 1.02

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