M Common HW #10 (LAST! x fi Dezto.mheducation.com/hm.tpx :: Apps P Pandora EMAIL

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M Common HW #10 (LAST! x fi Dezto.mheducation.com/hm.tpx :: Apps P Pandora EMAIL Th Moodle Top Hat whentowork N Netflix webadvisor BOM MCHEM Connect 9 CALC chem Li rent EJEMBE West Virginia Histor @Sign In: Discover Common Hw #10 (LAST!!) instructions I help Question 8 (of 16) Save & Exit Submit 8 value 6.25 points 1 out of 3 attempts Assistance Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3 CH2)3 (Kb = 5.2 x 10-4), with 0.1000 M HCl solution after the following additions of titrant. Check My Work Check Answer Report a Problem (a) 11.00 mL: (b) 20.90 mL: (c) 29.00 mL: 11:01 AM 11/30/2015

Explanation / Answer

(a) when 11.0 ml of 0.1 M HCl is added

moles of Et3N = 0.1 M x 0.02 L = 2 x 10^-3 mols

moles of HCl added = 0.1 x 0.011 L = 1.1 x 10^-3 mols

excess Et3N = 9 x 10^-4 mols

[Et3N] = 9 x 10^-4/0.031 = 0.029 M

[Et3NH+] = 1.1 x 10^-3/0.031 = 0.0355 M

pKb = -logKb = 3.284

pKa = 14 - pKb = 10.72

pH = pKa + log([base]/[acid])

     = 10.72 + log(0.029/0.0355)

     = 10.63

(b) When 20.90 ml of 0.1 M HCl added

excess moles of HCl = 0.1 M x 0.0009 L = 9 x 10^-5 moles

[H+] = 9 x 10^-5/0.0409 = 2.20 x 10^-3 M

pH = -log[H+] = 2.66

(c) When 29.00 ml of 0.1 M HCl was added

excess HCl = 0.1 x 0.009 = 9 x 10^-4 mols

[H+] = 9 x 10^-4/0.049 = 0.0184 M

pH = 1.736

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